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NOTES ON MECHANICS. 



PART I. - STATICS. 



NOTES ON MECHANICS. 



DESIGNED TO BE USED IX CONNECTION WITH EAK- 
KINE'S APPLIED MECHANICS. 



BY 

GAETAXO LANZA, S.B., C.E., 
« » 

ASSISTANT PROFESSOR OF MATHEMATICS AND MECHAIJICSy 

MASS. INSTITUTE OF TECHNOLOGY. 



PART L- STATICS, 









BOSTON: 

1872. 



Entered, according to Act of Congress, in the year 1872, 

By GAETANO LANZA, 
in the Office of the Librarian of Congress at Washington. 






NOTES ON MECHANICS. 



The fundamental idea of Mechanics is Motion. 

Force, as far as we have to consider it, may be defined as that 
which causes, or tends to cause, a body to change its state from rest 
to motion, or vice versa, or to change its motion as to direction or 
speed. 

Statics is that part of Mechanics which considers the relations 
of forces as producing pressures, or a tendency to motion. 

Dynamics considers forces as producing motion. 

Motion is a'fundamental idea ; it might be defined as change of 
position. 

Motion, as far as our senses are capable of recognizing it, is rela- 
tive. We cannot tell whether anything in the universe is abso- 
lutely at rest or not. 

All bodies on the earth partake of its motion, and hence even 
those that seem to be at rest when compared with the position of 
the earth, are really in motion. 

A person sleeping on a steamer is at rest with reference to the 
steamer, but in motion with reference to the earth. 

Hence we always assume some point as fixed; relatively to 
which we determine whether a body is in motion or at rest. 



4 LAWS OF MOTION. 

CONSIDERATION OF THE LAWS OF MOTION. 

There are three great laws of motion that lie at the basis of 
Mechanics. 

The proof of these laws lies in a proper conception of motion, 
and is derived from observation and experiment. 

FIRST LAW OF MOTION. 

A body at rest will remain at rest, and if in motion will continue 
to move uniformly and in a straight line, unless and until some ex- 
ternal force acts upon it. 

This means that matter is inert, that it has no power in itself to 
change its state from rest to motion, or from motion to rest. If it 
is at rest it remains at rest until some external cause puts it in mo- 
tion. If it is already in motion it continues to move uniformly, 
-and in a straight line, that is, with the same speed and in the same 
•direction ; and in order to bring it to rest, or to change its course 
or its speed, some external force is required. 

When we see terrestrial matter change its state from rest to mo- 
tion, or vice versa, we can generally assign the cause. 

A ball thrown from the hand soon comes to rest; the cause is 
Gravity and the resistance of the air ; if rolled along the ground 
Friction causes it to slacken its speed and soon come to rest ; if, 
however, we diminish the external force*s by rolling it on glass or 
on the ice, it continues its motion longer. 

Thus we see that the more we diminish the external forces the 
longer the motion continues, and hence we infer that were we able 
to remove them entirely the motion would continue forever. 

In the heavens the uniform motion of the planets shows us that 
the law is fulfilled there. 

SECOND LAW OF MOTION. 

If a body have two or more motions imparted to it simultaneously 
it will move so as to preserve them both. 



SECOND LAW OF MOTION. O 

That is to say, that every force that acts on a body has its full 
effect entirely independent of any other force or forces that may 
act on the body. 

Every relative motion is composed of the motion within a certain 
space, and also the motion of this space in reference to another 
space. Thus a person moving on the deck of a ship has two mo- 
tions in relation to the shore, viz., his own, and that of the ship. 

Suppose him to move in the direction of motion of the ship at 
the rate of 10 feet per second, while the ship moves at the rate of 
2b feet per second ; he appears to a person on the shore to move at 
the rate of 25 -f- 20 = 35 feet per second; if, on the other hand, 
lie walks in the opposite direction at the same rate, he appears to 
move at the rate of 25 — 10 = 15 feet per aecond. 

Suppose a body situated at A (Fig. 1) to have two motions im- 
parted to it simultaneously, one of which would carry it to B in one 
second, and the other would carry it to C in one second ; and we 
wish to determine where it will be at the end of one second, and 
what path it will have pursued. Imagine the body to move in obe- 
dience to the first alone during one second ; it would thus arrive at 
B ; then suppose the second motion to be imparted to it instead of 
the first, and it will arrive at the end of the next second at D, where 
BD is equal and parallel to AC. Now if the two motions are im- 
parted simultaneously instead of successively, the same point. D, 
will be reached in one second, instead of two ; and by dividing up 
AB and AC into the same (any) number of equal parts we can 
show that the body will always be situated at some point of the 
diagonal AD, or that it moves along AD. 

Hence the diagonal of a parallelogram constructed with two 
velocities, and with the angle inclosed by them, gives the direction 
and velocity of the resulting motion. 

A Force has three characteristics, which, when known, deter- 
mine it, viz., point of application, direction and magnitude. These 
ean all be represented by a straight line by making the length of 
the line proportional to the magnitude of the force ; hence a force 
can be represented in point of application, direction, and magnitude 
by a straight line. 



PARALLELOGRAM OF FORCES. 



Forces are most conveniently referred to our unit of weight as a 
unit ; they are evidently proportional to the amount of velocity or 
speed that they communicate to the same body. Thus a force 
which imparts to a body a velocity of 10 feet per second is twice 
as great as one that imparts to the same body a velocity of 5 feet 
per second. 

Hence from the parallelogram of motions we deduce that of 
forces, viz. : — 

If two forces acting at the same point be represented in point of 
application, direction and magnitude by two adjacent sides of a par- 
allelogram, their resultant will be represented by the diagonal of the 
parallelogram from their common point of application. 

Given two forces P and P' acting at the point A and inclined to 
each other at an angle 0, required to find the magnitude of the re- 
sultant force and its direction. 

Let AC (Fig. 1) represent P, AB represent P', and angle BAG 
= 6, then AD will represent in magnitude and direction the re- 
sultant force ; also let DAC = «, then from the triangle DAC we 
have 

AD 2 = AC 2 + CD 2 — 2 • AC • CD • cos DCA 
but DCA = 180° — and .-. cos DCA = — cos 
... W = P 2 + P /2 = 2PF cos or 
p v = V(P 2 + P /2 + 2PP' cos 0) 
This determines the magnitude of R,; and for its direction we 
have from the same triangle 

CD : AD = sin a : sin (180° — 0) or 
P'.: R = sin a : sin whence 

P' .'■ , P 

sm a = j5- sin ; so also sin (0 — a) = ^r sin 

EXAMPLE. 

1. Find the resultant of two forces equal to 10 and 17 respec- 
tively, and inclined to each other at an angle of 60°. 

Solution. R = V K 10 ) 2 + ( l7 ) 2 + 2-10-17 cos 60°^ =s 
V (100 + 289 + 170) -=23.643. sino = ^{f CT sin 60°. 






TRIAXGLE OF FORCES. / 

log 10 = 1.0000000 r R = 23.643 

log sin 60° = 9.9375306 Ans. ] a = 21° 29' 13" 

10.9375306 ( — a = 38° 30' 47" 

log 23.643 = 1.3737026 
log sin a = 9.5638280 
a = 21° 29' 13" = angle between the force 17 and resultant. 
2. Two forces equal to 47.34 and 75.46 respectively, act at an 
angle of 73° 14' 21" to each other, find the magnitude and direc- 
tion of the resultant. 

TRIANGLE OF FORCES. 

If we have the two forces OA and OB (Fig. 2) acting through 
one point O, their resultant is evidently OC ; and in order to bal- 
ance the two forces we shall have to apply a force equal in magni- 
tude and opposite in direction to OB. Now the sides of the 
triangle OAC represent in magnitude and in direction the force 
OA, the force AC and a force equal and opposite to OC, and these 
three forces, if applied at the same point, would balance each other, 
hence : — 

If three forces he represented in direction and magnitude by the 
three sides of a triangle taken in order, then these forces ichen 
applied at one point balance each other. 

Furthermore we have 

OC : OA : AC = sin OAC : sin OCA : sin COA, or 
R : : P : P' = sin 6 : sin (0 — a) : sin a. 

If in the value of R, page 6, the angle d becomes 90°, the re- 
sults become (Fig. 3) 



R = V (pa + P'2) ; s i n a = 



P' P 

R~ ;C03a== R 



EXAMPLE. 



Ex. Let P = 4, P' = 3, and then we have 

R = V (16 + 9) = 5, sin a = f , cos a = f 

log 3 = 0.4771213 

log 5 = 0.6989700 

log sin a = 9.7781513 a = 36° 52 f 11" 



DECOMPOSITION OF FORCES* 

DECOMPOSITION OF FORCES IN ONE PLANE, 

If, on the other hand, we have given a force R, making a given 
angle a with AC (Fig. 4) ; we can decompose the for^e H into two 
components, one in the direction AC, and the other perpendicular 
to it ; that is, we can find two forces acting in the directions AC 
and AB, respectively, that shall be equivalent to the single force R„ 
The two components of R are AC — P, and AB = P', and from 
the figure we derive 

P = R cos a ; P' = R sin a. 
Suppose now we have given three forces, P, P' and P", making 
angles a, a! and u! r (Fig. 4) respectively? with the line Ox (which 
we shall call the x axis), and we wish to determine the resultant of 
these three forces as to direction and magnitude. We first resolve 
each one into two components in the directions Ox and O?/, thus 
the components of P are OA and OA', of P', OB and OB 7 , of P", 
OC and OC, and moreover, 

OA = P cos a 7 OB == P' cos «', OC = P" cos «", 
OA 7 = P sin a, OB' =■ P' sin a', OC = P" sin a". 
Now these forces are together equivalent to a force 

P cos a 4~ P' cos a + P" cos a" in the direction Ox, 
and a force 

P sin a -\r P' sin a' + P" sin a! r in the direction Oy. 
The first may be represented by 2, P cos a, and "the second by 
2lV sin a; wheie 2J stands for "sum," either there are three or 
more forces, and now we can find the resultant of these two, viz., 
R =; J\(2 P cos af + (2 P sin a)% 

also a T7 the angle this resultant makes with Ox, is determined from 

the equation 

Z P cos a 
cos a T n 

EXAMPLES. 

1. Given P =-47.321 a == 21° 14' 27"- 

F = 73.423 a! = 43° 2' 54" 
p" = 43.214 a" = 82° 4' 12" 
F" = 23.457 a'" = 112° 41' 45" 
find the resultant force and its direction. 



EXAMPLES. \i 

Solution. P cos a = 44.1063 P sin a = 17.1439 

p' cos a = 53.6559 P' sin o! = 50.1196 

P" cos a!' = 5.9619 P" sin a" = 42.8007 

P>" cos a!" = 12.5404 P'" sin a'" = 29.9364 

vp cos a =91.1837 ^P sin a = 140.0006 

R = y<(91.1837) 2 + (140)*} = y(8314.4671 + 9600) 

= ^27914.4681 = 167.076. 
loff IPcosa = 2.1461295 



logR 






= 2.2229141 








log cos 


6 




= 9.9232154 







= 33° 4' 36". 


Given 


P 


= 


4.327 


a 


= 


77° 19' 47" 




P' 


= 


45.432 


o! 


== 


163° 58' 32" 




P" 


= 


105.27 


a" 


= 


275° 14' 12" 


Given P 


= 


57.432 


a 


— 


33° 4' 56" 




P 


— 


43.241 


r 

a 


= 


145° 10' 25" 




P" 


= 


12.556 


rr 
a 


= 


10° 15' 3" 




pw 


== 


U.doQ 


frt 

a 


•== 


190° 15' 3" 


Given 


p 


= 


123 


a 


= 


10° 2' 




P' 


= 


432 


o! 


= 


5° 4' 




P" 


= 


547 


o! f 


= 


48° 3' 




p'// 


= 


965 


a!" 


= 


72° 14' 




T>IV 


= 


127 


a lv 


= 


46° 37' 


Given 


P 


= 


.0123 


a 


= 


37° 




P' 


= 


.0047 


r 
a 


= 


42° 5' 17" 




P" 


= 


2.3456 


rr 
a 


= 


57° 3' 




p//7 


= 


9.0105 


rtr 

(A 


— 


42° 




piv 


= 


7.0003 


a IV 


= 


60° 




pv 


= 


8.001 


a v 


= 


54° 15' 


Given 


p 


= 


5 


a 


= 


cos" 1 f 




F 


== 


4 


r 
a 


= 







p/, 


= 


3 


rr 
a 


= 





6. 



7. Given three forces equal to 17, 15 and 13, respectively; the 
jingle between the first two is 73°, and between the second two 
47°, find the magnitude and direction of the resultant. 

8. Given four forces equal to 57, 8, 36 and 25, respectively; 



10 COMPOSITION OP FORCES. 

angle between the 1st and 2d is 60°, between 2d and 3d 30° ? be- 
tween 3d and 4th 45 °, find the resultant force and its direction. 

COMPOSITION OF FORCES IN SPACE. 

If we have several forces all acting at the same point, of which 
we wish to find the resultant, we assume first three lines, OX, OY 
and OZ (Fig. 5), at right angles to each other, whose common 
origin is the point at which the forces act. 

Now suppose OE = P to represent one of the forces ; we first 
resolve it into two forces, OC and OD, one of which acts along 
OZ, and the other in the plane XOY ; we then resolve OD into 
two components, OA and OB, acting in the directions OX and OY 
respectively. 

We thus obtain the three forces OA, OB and OC acting in the 
directions OX, OY and OZ respectively, which are together equiv- 
alent to the single force OE. These three components are the 
three edges of a rectangular parallelopiped of which OE = P is 
the diagonal. 

Let now a = angle EOX, ft = EOY, y = EOZ and we have 
from the right angled triangle EOA, EOB and EOC respectively 
the following equations : — 

OA = OE cos EOX = P cos a 

OB — OE cos EOY = P cos £ 

OC = OE cos EOZ = P cos y 
"We have also OA 2 +OB 2 = OD 2 and OD 2 +OC 2 = OE 2 
.-. OA 2 + OB 2 + OC 2 = OE 2 or P 2 cos 2 « + P 2 cos p + P 2 cos 2 y = P 
.\ cos 2 a + cos 2 j3 -\- cos 2 y = 1. 

When two of the angles, a, fj and y, are given, the third can be 
determined from the equation above. 

We proceed in the same way to resolve each of the forces into 
three components along the axes of x, y and z respectively, and 
we thus obtain a single force acting along the axis of x equal to 
1 P cos a = P cos a + P' cos a! + P" cos a" + &c, so likewise a 
force, 2 P cos £>, acting along the axis of y and 1 P cos y along 



COMPOSITION OF FORCES. 



11 



the axis of z. These three forces produce the same effect as all 
the forces with which we started. 

We next proceed to find a single resultant for these three 
forces. 

Let (Fig.6) OA = 2 P cos a* OB = 2 P cos (3. 
OC = 2 P cos y. 

Compounding OA and OB we find OD to be their resultant, 
and then by compounding OD and OC we find OE to be their re- 
sultant, and hence the resultant of the three forces OA, OB and 
OC. We have 

OE 2 = OD 2 + OC 2 — OA 2 + OB 2 + OC 2 or 
R 2 == (IP cos a) 2 + (IP cos (3) 2 + (IP cos 7 y 
I P COS a 



OA 

OE 

IP COS,? 



Also, cos a T = ^™ = 



R 



cos 



and likewise 
I P cos y 



This gives us the magnitude and direction of the resultant force. 





Examples. 




1 . Given. 






P =63.421 


a = 53° 15' (3 = 42° 17 . 




p' =49.357 


a' = 87° 3' 10" 


-/ = 72°3'27" 


P"= 2.729 


J 3"=70°5'17" 


r "=45°7'32" 


•P"' = 4.325 


a'"= 90° (S"' = 





Find the resultant, force and the angles it makes with the axes 
respectively. 



63.421 

49.357 

2.729 

4.325 



53° 15' 
87° 3' 10" 

90° 



ft 
42° 17' 

70° yi7" 

0° 



'2° 3' 27" 
45° V 32" 



COS a 

.59832 
.05141 
.62144 
.00000 



cos (8 

.73983 

.94997 

.34057 

1.00000 



COSy 

.30767 
.30807 
.70556 
,00000 



PCOS a 

37.94605 
2.53744 

1.69590 
0.00000 



42.17939 



P COS0 

46.92075 
46.88766 
0.92941 
4.32500 



99.06282 



(I P cos a) 2 = 1779,1009407721 
(IP cos |3) 2 = 9813.4423063524 
(IP cos y) 2 = 1342.7541538321 

R 2 = 12935.297400956G .\ R = 113,7334. 



P COS y 

19.51273 
15.20541 
1.92547 

0.00000 



36.64361 



12 CONDITIONS OF EQUILIBRIUM. 

log 2 P cos a = 1.6260003 log 2 P cos |3 = 1.9959100 log 2 P cos 7 = 1.56399S2 

logR = 2.0558879 log R = 2.0558879 log R = 2.0558879 



9. 


3701124 




9.9400221 


9.5081103 


« r = 68° 11' 2" & = 


= 29' 


5 25 


'27" 


Tx — 71° 12' 16" 


2. Given 












P = 4.324 


a = 47° 2' 




§ 


= 65° r 




P' =87.465 


a' =88° 3' 








/ = 10° 5' 


P" = 6.342 


a" = 68° 4' 




r 


= 83° 2' 




3. Given 












P = 5.107 


a =90° 




§ 


= 90° 




P' = 7.321 


a = 0° 










P" = 4.325 






p 


= 0° 




p'" = 74.215 


a'" = 73° 2' 


15" 






r "'=r45°17' 3" 


4. Given 












P =43.217 


a = 30° 




'? 


= 60° 




P' =59.017 


a =45° 








y = 45° 


P" = 63 


a =72° 3" 


14" 






r =43° 2' 29" 


P'" = 42 






P 


= 83° 7' 


r =54° 5' 


5. Given 












P = 87 


a = 30° 




(8 = 


= 60° 




P' = 42 


a = 45° 




P' = 


= 45° 




P" =112 


a = 90° 




(3 = 


= 90° 




P'" = 3 


a = 90° 




P = 


= 30° 




P /T = 85 


a = 15° 




tr 


= 105° 





Find in each case the magnitude and direction of the resultant. 

CONDITIONS OF EQUILIBRIUM. 

We have seen that in the composition of forces we begin by re- 
solving each force into three components along three rectangular 
axes ; and for the final resultant force we have 

R 2 = (I P cos a) 2 + (IF cos $y + (2 P cos r )\ 
Now if the forces balance each other, or are in equilibrium, the 
resultant must be equal to 0, that is, R — .*. W = 

.-. (2 P cos «) 2 + (Z P cos /5) 2 + (2P cos 7 y = 0. 
But this expression is the sum of three squares, none of which 



RIGID BODIES. 



13 



can be negative ; hence each one of the quantities in the brackets 
must be separately equal to 0, hence the conditions for equilibrium 

are 

2 p cos a = ; Z P cos ,3 = ; and Z P cos y = 0. 

When the forces are all in the same plane, that of (x, y), cos y 
= 0, since y = 90° in every case, and hence 

cos 2 a _j_ cos 2 p _ i^ or cos p — s i n a , hence the conditions of 

equilibrium become in this case 

2 p C os a = ; Z P sin a =■ 0, 

a result which might have been obtained from the fact that for 
forces in one plane R 2 = (Z P cos «) 2 + (2 P sin a) 2 , and hence 
when E = 0, 2 P cos a = and 2;Psin a = 0. This shows that 
in order that a set of forces may be in equilibrium the sums of 
their components along each of the axes must be separately equal 
to 0, 

STATICS OF RIGID BODIES. 

Thus far we have always conceived our forces^ to act upon a sin- 
gle point, and have not considered the case when we have forces 
acting at different points of a body. 

A Rigid Body is one that does not undergo any alteration of 
shape, when external forces are applied to it. Strictly speaking 
no body is absolutely rigid. 

If a force be applied to a rigid body (Fig. 7), say at the point 
A, it communicates motion to this particle ; but since the body 
does not alter its shape the next particle must move the same 
amount, and in the same line, and so the next, so that each parti- 
cle in the line of the force moves just as though the force were ap- 
plied there. 

And moreover if we apply an equal and opposite force at any 
other point of the line of action of the first, as B, it produces 
equilibrium, hence in the case of a rigid body the point of appli- 
cation of a force may be transferred to any other point in the line 
of action of the force without changing the effect. 



14 MOMENT OF A FORCE. 

COMPOSITION OF TWO FORCES ACTING AT DIFFERENT 
POINTS OF A RIGID BODY, 

Suppose the force P (Fig. 8) to be applied at the point A, and 
P' at B. Now, since the point of application of a force may be 
moved anywhere in its line of action without changing the effect, 
we produce the lines representing the directions of the forces till 
they meet in O, and suppose P and P' both to act at O. Now we 
find by forming the parallelogram ODFE the resultant of these 
two forces to be R — OF acting in the direction OF : we may im- 
agine its point of application to be anywhere in its line of action, 
say at C. 

MOMENT OF A FORCE. 

We have hitherto considered forces so far as they tend to pro- 
duce motion in a straight line, that is motion of translation. 

But suppose the point A of a rigid body (Fig. 9) to be fixed, 
and a force to act in the direction, BD ; draw AC perpendicular to 
BD and transfer the point of application of the force to C, which 
does not alter the effect ; now the point A being fixed, the only 
motion that C can have is a motion of rotation around A as a cen- 
tre. 

It becomes necessary for us to find a measure for the rotatory 
effect of a force in regard to any point of a body. 

Let, as before (Fig. 10), the forces P act at A, and P' at B, and 
suppose on finding the resultant we find it to be CD. If now we 
apply at C a force R, equal and opposite to CD, this will, together 
with P and P', keep the body in equilibrium. Now let us examine 
the effects of the respective forces P, P- and R in regard to rota- 
tion around C, since we may regard C as a fixed point, the body 
being at rest. 

It is evident that the force P tends to make the body rotate in 
one direction, and the force P' tends to make it rotate in the oppo- 
site direction, and since the body has no motion of rotation, and 
these two are the only forces that have a tendency to give it such 






STATICAL COUPLES. 15 

a motion, therefore the rotatory effect of these two forces around C 
is balanced, and the question arises how to find the condition of 
such balancing. From C draw the two perpendiculars, p and p\ 
on the directions of the forces, and we have the area of the trian- 
gle OCG equal to | p • OG ; and that of OCH equal to \p • OH; 

but since these areas are equal p OH = p' OG, or — , = Tyrj. but 

OG P' p P' _, 

OH = P •>' = P> m P*-P r 

that is, the product of the force P by the perpendicular from C on 
its direction is equal to the product of P' by the perpendicular 
from C on its direction, and this is the necessary and sufficient con- 
dition that the two forces, P and P', may produce no rotation 
around C, bearing in mind, of course, that the two tendencies of 
rotation must be in opposite directions. 

Definition. The product of a force by its perpendicular dis- 
tance from a point, is called the moment of the force with respect 
to the point. 

Since two forces that have the same moment with respect to a 
point produce the same rotatory effect, therefore we may take the 
moment of a force as the measure of its rotatory effect. 

STATICAL COUPLES. 

Suppose (Fig. 11) we have two equal forces acting on a body 
in parallel but opposite directions ; draw the line AB perpendicu- 
lar to both, and transfer their points of application to A and B re- 
spectively. Xow it is evident that such a pair of forces can have 
no single resultant ; for such a resultant would produce motion in 
a definite direction, and anv reason that could be given for assion- 
iDg the motion with respect to one of the forces, would be equally 
valid with respect to the other, hence the only effect of such a pair 
of forces is a motion of rotation. 

Such a pair of forces is. called a Statical Couple. 

Its effect is evidently to produce rotation around an axis perpen- 
dicular to the plane of the couple. 



16 STATICAL COUPLES. 

This rotation may be in the direction of the hands of a watch, 
when it is called right-handed, and is accounted positive, or in the 
opposite direction, when it is left-handed, and is accounted nega- 
tive. 

The rotation produced by a couple is the same, no mat- 
ter at what point we take the axis, provided only this axis be per- 
pendicular to the plane of the couple ; and the rotatory effect is 
always measured by the product of the force at either end multi- 
plied by the perpendiadar distance between them, which is called 
the arm of the couple, and the product is called the Moment of the 
Couple. 

Let the couple (Fig. 12) be that with arm AB and with force 
P, and let us take O as the fixed point ; draw from O a perpendic- 
ular to the forces, and transfer the point of application of P to D, 
and of — P to C. Now the rotatory effect of P is POD, and that 
of - — P is POC ; or the resultant rotatory effect is the di; erence 
between these two, since they act in opposite directions, or POD 
= P-AB. 

IP now the arm of the couple be turned through any angle, it 
changes neither the direction of rotation nor the rotatory effect ; 
and the same is the case if the couple be situated in my plane 
parallel to the plane where it now is. 

Hence two couples are equivalent whose planes are coincident 
or parallel, whose moments are equal, and whose s-ense (direction 
of rotation) is the same. 

COMPOSITION OF STATICAL COUPLES. 

First suppose the couples to be situated in parallel planes, or in 
the same plane. If their forces are not the same reduce them to 
the same force, and then place their arms in the same straight line, 
as in the figure ; the forces P and — P at B evidently balance each 
other, and may be removed ; and we thus have left the couple 
whose arm is AC = AB -f- BC, that is, the sum of the two arms. 
Multiply through by P, and we have P-AC = P-AB + P-BC, 
which shows that 



MOMENT OF A COUPLE* 



17 



The moment of the couple, which is the resultant of ttvo or 
more couples in the same or parallel planes, is equal to the sum of 
the moments of the component couples, 

EXAMPLES. 

1. Convert a couple whose force is 5, and arm 6, to an equiva- 
lent couple whose arm is 3. Find the resultant of this, and an- 
other couple in the same plane, whose force is 7, and arm 8 ; and 
find the force of the resultant couple corresponding to an arm 5. 

Solution. Moment of 1st couple = 5 X 6 = 30. 

When arm is 3 force = 3 3° = 10. 

Moment of 2d couple = 7 x 8 = 56. 

Moment of resultant couple = 30 + 5G = 86. 

When arm is 5 force = % & = 17 } T . 

Arm. 



2. 


Force. 


Arm. 




Fore 




r 12 


17 




5 




3 


8 






Given the fol- 


5 


7 


Convert to 


8 


lowing couples in *< 


6 


9 


couples having < 


6 


one plane. 


12 


12 


the following. 






10 


9 




4 




- 14 


6 




I 



20 

The first and the last three are right handed, the second, third 
and fourth are left handed ; find the moment of the resultant 
couple, and also its force when it has an arm 11. 

Arm, 



3. 


Force. 


Arm. 




Force-* 




7 


9 




- n 


| 


14 


12 






Given the fol- 


15 


9 


Convert to 


6 


lowing couples 
all in the same 


12 

6 


15 

9 


couples having 
the following. 


43 


plane. 


3 
1 


2 
1 




12 




1 


i 

4 




9 



30 



12 



The first four are right handed, last four left handed ; find result- 
ant couple, and its arm when the force is 25. 



18 



COMPOSITION OF COUPLES. 



Force. 


Arm. 


10 


12 


4 


3 


6 


9 


10 


9 


5 


12 


12 


3 



Force. 


Arm. 


r 8 






7 


6 






5 


9 






4 


ift handed : 


find 



The first three are right handed, and last three left handed ; 
resultant couple. 



REPRESENTATION OF A COUPLE BY A LINE. 

From the preceding we see that a couple remains always the 
same as long as its moment does not change ; the direction of its 
axis (t\ e., a line drawn perpendicular to the plane of the couple), 
and the direction in which it tends to make the body turn remains 
the same. 

Hence a couple may be represented by drawing a line in the di- 
rection of its axis, and laying off a distance representing the mo- 
ment of the couple on this axis. 



COMPOSITION OF COUPLES SITUATED IN PLANES IN- 
CLINED TO EACH OTHER. 

Suppose we have two couples situated neither in the same plane 
nor in parallel planes, and we wish to find their single resultant 
couple. It is first necessary, if their arms are not equal, to sub- 
stitute for them equivalent couples having equal arms ; then transfer 
them in their own planes respectively to a position such that their 
arms shall lie in the line of intersection of the two planes and 
make their arms coincide. 

Suppose after this transformation has been affected 00' (Fig. 14) 
to be the common arm, P and — P the forces of one couple, P' and 
— P' those of the other. The forces P and P' have for their result- 
ant R. 



COMPOSITION OF COUPLES. 19 

Now OP and OP' are both perpendicular to OO', hence their 
plane and consequently the line OR is perpendicular to OO r . — P 
and — P' give a resultant — R perpendicular also to OO' and par- 
allel to R. The resultant of the two couples is consequently a 
couple with the same arm, and with a force R, the diagonal of the 
parallelogram on OP and OP', so that 

R = J (p2 + P'2 + 2PP' COS 0) 

where is the angle between the planes of the couples. 

Now if we draw at the line Oa perpendicular to 00' and 
also to OP, consequently perpendicular to the plane of the first 
couple ; and Ob perpendicular to the plane of the second ; and 
make the lengths Oa and 05 proportional to P and P' so that 

Oa:Oh :: P : P' :: 00' • P : 00' • P' : 
first we observe that these lines Oa and Ob lie in the same plane 
with OP and OP', perpendicular to 00'; and moreover if we con- 
struct on Oa and Ob a parallelogram the diagonal Oc will be per- 
pendicular to OR, and 

Oc : OR : : Oa : OP : : Ob : OF 
since the angle aOb = POP' ; hence if Oa and Ob represent the 
moments of the couples (of force) P and P' respectively, Oc will 
represent the moment of the resultant couple whose force is R, 
and 

Oc = y(Oa 2 -t- Oh 2 + 20a • Ob cos 0) 
let Oa = L, Ob = M, we have OC = Gr, and, 
G = y (L 2 + M 2 + 2LM cos 0). 
We therefore see that we can compound and resolve couples just 
as we do forces, provided we represent a couple by its moment 
axis. 

EXAMPLES, 

1. Two couples act in planes inclined to each other at an angle 
of 65° 5'; their moments are 43 and 75 respectively, and both are 
right handed when looking along their axes. Find the moment of 
the resultant couple and the angle of inclination of its axis to that 
of the first. 



20 



EXAMPLES. 



Solution. 6 = 65° 5', L = 43, M = 75. 

G = y(L 2 + M 2 + 2LM cos 0) = 
y \ (43) 2 + (75) 2 + 2-43-75 cos 65° 5^=^10191.3850 = 100.9523. 
Now for the angle its axis makes with that of the first, 
sin a = T oi%5 2 sin 65 ° 5 '- lo g sin 65 ° 5 ' = 9.9575697 



log 75 



log 100.952 



= 42° 21' 33' 



: 1.8750 613 
1.8326310 

= 2.0041149 
9.8285161 
_ a — 22° 43' 26". 



If one of the couples were left handed w hen looking along its 
axis as given, we should only look at it from the other end of the 
axis and the only change made would be 

= 180° — 65° S', or = 114° 55' .-. 
G = ^(43) 2 + (75) 2 +2 -43 • 75 cos 1U° 5o'\ = 
</(4756-6150) = 68.9682. 
log sin 114° 55' = 9.9575697 
log 75 = 1.8750613 



log 68.9682 
log sin a 



1.8326310 

= 1.8386488 
= 9.9939822 
a = 80° 29' 1". 



34° 25' 59". 



2. Given two couples whose axes are inclined to each other at 
an angle 37° 15' and whose moments are 83 and 112 respectively; 
both right handed. 

3. Given L = 57.32, M = 243.12, = 170° 3', find G and a. 



4. 


Given 


L = 125.34 


M= 43.27 


= 85° 3' 


5. 


u 


L= 43.29 


M= 32.75 


= 63° 15' 


6. 


u 


L= 95.32 


M = 142.31 


0= 27° 2' 


7. 


u 


L= 86.41 


M= 63,42 


0= 33° 5' 


8. 


a 


L= 93.75 


M= 42.36 


= 87° 32' 


9. 


u 


L = 532.01 


M = 345.07 


= 115° 25' 


10. 


u 


L == 327.05 


M = 315.04 


= 135° 13' 



PARALLEL FORCES. 



21 



11. « L = 0.012 M= .234 = 60° 

12. " L = 103.42 M = 79.63 => 32° 15' 
Find G unci a in each of the above. 

PAEALLEL FORCES. 

If we have two couples (Fig. 15) of equal moment in the same 
plane one of which is right handed, and the other left-handed, 
these two couples, if there be no other force acting, will evidently 
neutralize each other's effect and produce equilibrium. Let us now 
place them with their arms in the same line and contiguous to each 
other as in the figure ; P is the force of one and a its arm, P' is the 
force of the other and b its arm, and Pa = P'J. 

The couples thus placed will form a system of three parallel 
forces which are in equilibrium, and hence the force P -\- P' at O 
balances the two forces P at A and P' at B, hence the resultant of 
P at A and P' at B is a force equal and opposite to P +P' at O ; 
that is in the same direction as the forces. Now O is such a point 

that 

P P' P + P' 

Pa = P'6. or - - = — , and this — j — r " 

o a a -f- o 

The preceding shows us that the resultant of two parallel forces 
is a force parallel to the two, and equal in magnitude to their sum ; 
and secondly that it acts at a point along the line joining the point 
of application of the two, which divides this line inversely as the 
forces ; that is the point of application of the resultant is on the 
line joining the points of application of the components and nearer 
the larger force. This is nothing more than the principle of the 
lever. If a man wishes to raise a weight with a straight lever the 
power he exerts multiplied by the distance from his hand to the 
fulcrum is equal to the weight lifted multiplied by the distance 
between the weight and the fulcrum. 

Suppose we have two forces (Fig. 16), P at A and P' at B, act- 
ing in opposite directions, of which we wish to find the resultant. 
If we apply at a force equal to P — P' acting in the direction of 
the less force, being such a point that P ■ OA = P' • OB we have 



22 PARALLEL FORCES. 

evidently equilibrium, hence the resultant of the two forces P and 
P r acting in opposite directions would be a force equal to P — P' 

P P' 

acting at a point such that P ■ OA = P' • OB, or Tyrr = ftt 

where AB is divided externally in the ratio P : P'. Moreover 
P — P' is the algebraic sum of the two forces above. 

EXAMPLES. 

1. Suppose we wish to make a balance such that one ounce at 
the end of the long arm shall balance one pound at the end of the 
short arm, the weights of the two arms being already inversely as 
their lengths. What must be the length of each arm, supposing 
the length of the beam to be 2 feet ? 

Solution. Let x be the length of the long arm, 2 — ■ x will be 
the length of the short arm ; then 1 oz. : 16 oz = 2 — ■ x : x .\ x 
= 32 — 16x y or x — f| ft., and the short arm = T 2 T ft. 

2. Suppose in a similar balance, whose beam is 35 in. long, 
that 1 oz. is to balance 12. 

3. Beam = 28 in. 3 oz. to balance 15. 

4. Beam = 8 v ft. 5 oz. " « 25. 

5. Beam = 25 ft. 4 oz. " " 4 oz. 

In the previous reasoning the forces have been so taken that the 
line connecting their points of application was perpendicular to 
them, but we might have assumed it oblique and we should have 
the same results ; for, from what we have seen, the point of appli- 
cation of a force may be shifted to any point whatsoever in its line 
of action. 

For the remainder of Section 3d of Rankine's App. Mech., 
I would refer the student to the text, and I shall only give exam- 
ples and some comments on Art. 47. 

If we assume three rectangular coordinate axes OX, OY and 
OZ (Fig. 17), and we have a couple (force P, arm OA) in the 
plane ZOX, the axis of this couple will be perpendicular to this 
plane or lie in the axis OY. The couple in the figure being right 
handed we lay off on OY, a distance OB representing the moment 



PARALLEL FORCES. 23 

of the couple ; and in the positive direction and we have a full 
representative of the couple. 

So if we have replaced all the forces by equal forces acting at O 
in the line OZ, and the necessary couples acting in the planes 
ZOX and ZOY respectively, we have for the resultant of the 
forces acting in OZ, R = I F. If now this should be equal to O, 
that is if the sum of the forces acting upward should be equal to 
the sum of those acting downward, then we have no single force 
but two couples, 2, xF and H yF acting in the planes ZOX and 
ZOY respectively. Now 2J xF will be represented (Fig. 18), by 
a line OB = M y = 2 xF laid off on the y axis and 2 xF by 
OA = M x = J£ yF laid off on the x axis. Compounding these 
two couples we obtain a resultant couple whose moment 
M r = J (OA 2 + OB 2 ) = V (M x 2 + M y 2 ) and which acts in a plane 
perpendicular to OC. 

If 6 is the angle its axis OC makes with OX cos = -^r^, and 

this is also the magnitude of the angle which its plane makes with 
the plane ZOY. 

EXAMPLES. 

1. With solution. Given the following forces, all parallel to 
OZ, x and y are their points of application. 



R = ^F = 95. 

ZxF= 1127;2yF== 1165. 

-r> =1127 1182 

X r 95 X1 ¥5' 

y = 11 ft 5 = 1-> 5 - 

95 1127 1165 

We thus have a force 95 acting at a point whose coordinates are 
llff, and 12 T %. We might also consider the force as acting at 
the origin, and then we should have in addition the two couples 
2 xF and 2 yF acting in the planes ZOX and ZOY respectively. 
We may now compound these two couples and obtain a couple. 



F 


X 


y 


xF 


yF 


63 


10 


21 


630 


1323 


14 


35 


— 6 


490 


— 84 


7 


— 8 


— 26 


— dQ 


— 182 


3 


45 


4 


135 


12 


8 


— 9 


12 


— 72 


96 



24 PARALLEL FORCES. 

M r = x/(M x 2 + M y 2 ) = y^(1127) 2 + (1165) 2 ^ = 1620.9 + 

M x 

whose axis makes with OX an angle 6, such that cos = -jrp = 

tWoV log 1127 =3.0519239 

log 1620.9 = 3.2097562 

log cos = 9.8421677 = 44° 3' 2". 
This couple and the force 95 at the origin are equivalent to a 
parallel and equal force at the point lift, 12 T %. 



2. 


F 


X 


y 


3. F 


X 


V 


4. F 


X 


# 




12.3 


4 


7 


14,7 


5 


6 


10 


12 


13 




6.2 


— 5 


8 


11.3 


— 2 


— 3 


5 


6 


7 




12.5 


4 


— 10 


17.7 


5 


7 


9 


14 


15 




8.2 


7 


— 5 


18.9 


4 


3 


11 


28 


53 




6.3 


4 


— 9 


12.3 


8 


— 9 


49 


3 


2 


5. 


F 


X 


y 


6. F 


x 


2/ 


7. F 


X 


# 




10 


7 


3 


7 


5 


1 


8 


4 


8 




-8 


6 


4 


3 


4 


3 


— 5 


3 


6 




6 


3 


2 


5 


7 


9 


— 2 


2 


5 




— 9 


5 


7 


— 8 


12 


15 


— 3 


1 


4 




1 


2 


i 

3 


— 7 


17 


16 


■ — 5 


7 


9 


8. 


F 


X 


2/ 


9. F 


X 


y io 


. F a 




y 




9.2 


8.4 


12.3 


127.2 


12 .4 


9.4 


10.2 7 


3 


5.4 




7.1 


5.6 


8.2 


835.7 


6.37 


8.3 


12.7 21.4 


9 




9 


12.7 


9.8 


946.8 


4.37 


7.2 


7 5 




6 




6.3 


4.2 - 


-7.1 


832.3 


8.73 


9.1 


8 9 




10 




8.3 - 


-7.1 


8.3 


946.7 


7.65 


10.8 


11 12 




13 




9.5 


43.1 


72.6 


832.8 


8.51 


9.3 


7.4 8.2 


9 



Find the resultant force and the coordinate of its point of appli- 
cation in each of the above. 



CENTRES OF PARALLEL FORCES. 25 

CENTRES OF PARALLEL FORCES. 

We have seen that in combining any number of parallel forces 
we obtain, (except when X F = O), a single resultant acting in a 
definite direction whose point of application may be taken any- 
where along its line of action. 

If we have two parallel forces (Fig. 19). P and P' applied at 
the points A and B respectively of a rigid body, from what we 
have already seen, the resultant is a force P -|- V applied at the 
point C which divides AB inversely as the forces ; now the point 
of application may be assumed at C, or any other point of CR as 
long as the forces remain in the same position, but if we suppose 
them turned around and taking the positions AO and AO' the 
resultant will still pass through C, but no other point of the line 
CR, hence C does not change as long as the points of application 
A and B of the forces remain the same, and their relative magni- 
tude does not change, and they remain parallel, no matter how 
they are turned. This point C is called the centre of the two 
parallel forces ; and in general 

The Centre of a System of Parallel Forces is the 
point of application of the resultant of the system, however the 
system be turned around, provided only the points of application 
of the single forces remain the same, and their directions continue 
parallel to each other. 

Hence; in finding the centre of a set of parallel forces, we may 
suppose the forces turned around through any angle whatever, and 
the centre will remain the same. 1 

EXAMPLES. 

Given the following parallel forces and the coordinates of their 
points of application, find the centre of each system. 



F 


X 


y 


z 


2. F 


X 


y 


z 


75.3 


4 


9 


8 


82.6 


6 


13 


48 


83/2 


—6 


5 


7 


73.1 


7 


25 


05 


—9 


6 


3 


—8 


54.G 


— y 


3.7 


99 


12 


5 


— 7 


3 


33.7 


12 


42 


87 


4.7 


5.3 


—8.1 


9.7 


57.5 


—15 


31 


78 



x \ now refer the studeut to Rankine's App. Mech., Section 4, for the determination 
of the coordinates of the centre of a set of parallel forces. 



26 FORCES IN A PLANE. 



F 


X 


y 


z 


4. F 


X 


y 


z 


63 


87 


21 


1 


1 


1 


i 


1 


48 


93 


7 


i 

2 


1 


—1 


— i 


— 1 


74 


64 


6 


2 


1 


—1 


i 


— 1 


82 


82 


3 


1 


1 


1 


— i 


1 


91 


97 


4 


1 


—1 


1 


i 


1 



INCLINED F0ECE8 AT DIFFEKENT POINTS ALL ACTING 
IN THE SAME PLANE. 

Another mode of demonstrating Rankine's Art. 59. 

Let CF, (Fig. 20), be one of the forces. We first decompose 
it into the two forces CD and CE acting in the directions OX and 
OY respectively, CD = F cos a, CE = F sin a. Now transfer the 
point of application of CD to B ? and of CE to A ; this does not 
alter their effect. 2d. Apply two forces at O in opposite direc- 
tions, in the line OY, each equal to CE == F sin a ; and two in the 
direction OX, each equal to F cos a. They do not alter the effect 
at all. 

We have now, instead of the force FC, six forces of which AH 
and OG form a couple (right handed in the figure), and BL and 
OK a left handed couple. The moment of the first is AH ■ AO 
= F sin a-x, and of the second BL ■ OB — F cos a-y {x and y 
being the coordinates of C) and since these two couples are in the 
same plane, the moment of their resultant couple will be Fx sin a 
— Fy cos a = F (x sin a — y cos a). We have also a force OM 
= F cos a, acting along OX and 0~N = F sin a along OY. 

We decompose in the same manner each of the forces into 
two, F cos a along OX, and F sin a along OY, and also a couple 
F(x sin a — y cos a). 

The resultant force along OX will be 2 F cos «, along OY 
2 F sin a, and the moment of the resultant couple 
M = 2 F (x sin a — y cos a). 

The two forces, 2 F cos a and 2 F sin a, give a resultant 
R = J \{2 F cos ay + (2' F sin af\ 



EXAMPLES. 27 

which we determine as in Art. 12 of Rankine. We may then 
determine the point of application of the single force equivalent to 
this resultant and the couple by Art. 12 of Rankine, or by the 
formula 

£ F.x sin a I F y cos a 

X ~ Z F sin a ' y ~~ I F cos a ' 

the same that have been developed for parallel forces. 









EXAMPLE. 








1. Solution 


. 












F x y a 


£OS a 


sin a 


FcOS a 


F sin a 


F x sin a 


F y cos a 


5 % 2 32° 


,84805 


.52992 


4.24025 


2.64960 


7^4880 


8.48059 


10 1 3 43^ 


,73135 


.68200 


7.31350 


6.82000 


6.82000 


21.94050 


7 4 2 54^10' 


.58543 


,81072 


4.09801 


5.67504 


22.70016 


8,19602 


--3 7 9 59° V 


.51329 


,85821 


—1.53987 


—2.57453 


—18.02241 


—13.85883 


2 6 1 -35° & 


.81865 


.57429 


1.63730 


1.14858 


6.89148 


1.63739 


1 4 2 41° 


.75471 


,65606 


0.75471 


0.65606 


2.62424 


1.50942 




16.50390 


14.37465 


28.96237 


27.80491 












27.90491 














1.0o746 = 


Moment of 












result, couple. 


(-£■ F cos 


af = 


272.3787152100 








(2"F sin 


af = 


206.6305626225 









479.0092778325 

R ~ y |(IF cos «) 2 + (JF sin a) 2 \ = 21.88628 
log 16.5039 = 1.2175865 
log 11.8862 = 2.3401718 
log cos a T — 9.8774147 

«j = m* 3' 19" 

3. F x y a 

11 9 8 210° 4' 

74 \ 1 54° 2' 

63 i — | 73° 12' 

16 \ \ 87° 43' 

25 I ^ 93° 57' 

33 T \ 4 82° 9' 



F 


X 


y 


a 


12 


27 


33 


15° 3' 


4 


32 


45 


75° 2' 


8 


—5 


—7 


63° 5' 


9 


4 


—3 


42° 9' 


21 


1 


2 


110° 4' 


45 


i 


— 1 


163° 2' 



F 


* 


y 


a 


10 


3 


2 


30°' 


5 


7 


— 1 


45° 


3 


—2 


1 


72° 


7 


1 


4 


125° 


3 


4 


—1 


75° 



28 FORCES IN SPACE, 

4. F o& y a 5. 

4.2 18 27 14° 29' 

3.5 2 4 53* 37' 
5.7 9 8 82° 51' 

8.6 10 12 97° 50' 
9.3 20 5 81° 14' 

11.2 5 8 93° 17' 

27.5 2 7 80° 4' 

6. 5 7 9 60* 7. 10 5 17 30° 

3 4 10 30° 5 9 82 45 a 

3 8 11 150° —3 11 91 60 c> 
—2 9 2 0° —2 23 83 90° 
—13 1 60° 1 25 112 0° 

8. 3 2 5 75° 9. 7.1 5 3 150* 

4 7 2 45° 8.2 9 —2 210° 
—7 6 3 90° —6.3 3 —4 60° 
—2 1 ™4 180° 9.4 —2 5 15 a 

2 4 10° --12.5' 1 3 87 a 

Find in each of the above the resultant force, and couple. 

COMPOSITION OF FORCES IN SPACE. 

Suppose we have a number of forces applied at different points 
of a body, and in different directions, of which we wish to find 
the resultant ; we refer them all to a set of three rectangular axes* 

Suppose PR (Fig. 21) to be one of the forces == F. First we 
draw through P parallels to OX, OY and OZ, and form a rectan- 
gular parallelopiped ; we thus obtain the three forces PK, PH and 
PG, which are equivalent to PR. Now let 

a = RPK, (3 = RPH, and y = RPG ; 
also let (x = OA, y = OB, z = OC) be the coordinates of P. 

We now can transfer the point of application of PK to E with- 
out altering the effect, and we thus have EL instead of PK. Now 
introduce at B and at O two forces opposite in direction, each 
equal to EL, and parallel to EL; these four forces do not alter 



FORCES IN SPACE- 29 



EL, BM, BN, OS and OT. The two forces EL and BN form a 
couple, parallel to the zx plane whose axis lies in the Y axis, the 
moment of which is EL-EB, but EL = F cos a and EB = 2:, 
hence moment = F^ cos a : the two forces BM and OT form a 
couple perpendicular to the z axis whose moment is BM • OB = 
F y cos a. Now do the same for the other two forces PH and 
PG, and we shall finally have instead of the force PR three forces, 
F cos «, F cos f>, F cos 7, acting at O in the directions OX, OY, 
OZ respectively ; together with six couples, two of which are in 
the zx plane, two in the zy plane, and two in the xy plane. They 
thus form three couples whose moments are as follows : — 
Around OX F (y cos 7 — z cos /3) 
" OY F (z cos a — x cos 7) 
" OZ F (x, cos £> — y cos «) 
Treat each force in the same way, and we shall have in place of 
all the forces three forces acting at O ; 2 F cos a along OX 5 
-T F cos (3 along OY, and J£F cos y along OZ ; also three couples 
whose moments are as follows : — 

Around OX M 1 = 2 ^F (y cos 7 — z cos £>) \ 
OY M 2 = 2: \F (z cos a — .rcos y)\ 
« OZ M 3 = 2 \ F (x cos p — y cos a) \ 

The three forces give a resultant 
R = j \ (2 F cos ay + (J£ F cos p°) 2 + {2 F cos yf\ 

2 F cos a 2 F cos S 2 F cos 7 
cos a r — g ; cos p r = j> •; COS 7, =* ^ 

as in Art. 60 of Rankine. 
Bear in mind also that 



cos" a 



+ cos 2 p> + cos 2 7 — 1 



The three couples give a resultant 

M — y(M x 2 + M 2 2 + M 3 2 ) 

M, M 9 M, 

cos I — jj ; cos ix = ^p cosv= M 

/, /Jt and v being the angles the axis of the resultant couple makes 
withOX, OY and OZ respectively. 



30 



FORCES IN SPACE. 





Example 1 


, with 


Solution 






F 


X 


y 


z 


a 


P 


7 


COS a 


COS |3 


12 
13 


— 14 
1 
9 
2 


13 

i 


2 

3 




47° 
15° 


49° 
32° 

; 47° 


74° 
82° 
43° 


.68200 
.45258 
.96593 
.00000 


.65606 
.84805 
.21819 
.68200 



.32320 
.27564 
.13917 
.73135 



8.18400 
5.88354 
9.65930 
0.00000 



7.87272 

11.02465 

2.18190 

2.04600 



3.87840 
3.58332 
1.39170 
2.19405 



23.72684 23.12527 11.04747 



Fy cos a 


Fz cos j8 


¥z COS a 


¥x cos y 


Fx cos |3 


Fy cos a 


50.41920 


15.74544 


16.36800 


— 54.29760 


— 110.21808 


106.39200 


— 1.79166 


33.07395 


17.65062 


3.58332 


11.02465 


— 2.94177 


11.13360 


15.27330 


67.61510 


12.52530 


19.63710 


77.27440 


2.19405 


0.00000 


0.00000 


4.38810 


4.09200 


0.00000 


61.95519 


64.09269 


101.63372 


— 33.80088 


— ,75. 46433 


180.72463 


64.09269 




— 33.80088 




180.72463 




— 2.13750 




135.43460 





— 256.18896 






Ei = 23.72684 


M 1 = 


— 2.13750. 






E 2 = 23.12527 


M 2 = 


135.43460. 






R, = 11.04747 


M 3 = — 


256.18896. 





R= y(Rj!> + R/ + R 3 2 ) = y (1219.7876423594) — 34.92545. 

M = y(M x 2 + M 2 2 + M 8 2 ) = y (83979.89300929) = 289.7928. 



logRx 
logR 
los COS a 


= 1.3752398 
= 1.5431419 

r = 9.8320979 


logR 2 
logR 

log COS £r = 


= 1.3640868 
= 1.5431419 
= 9.8209449 




logR 3 
logR 
log cos 


*= 1.0432628 

- 1.5431419 

yr = 9.5001209 


«r — 


47° U 


>/ 24" (3, =* 48° 


32' 15" 


Yt 


== 71 c 


D 33' 36". 


log M A 
log M 

log cos (180° 


-A) = 


0.3299061 
2.4620876 

7.8678185 


log M 2 = 
log M = 

log COS IX = 


2.1317296 

2.4620876 
= 9.6696420 


log M 3 
log M 
log cos (180 c 


= 2.4085603 

= 2.4620876 

> — v) = 9.9464727 


A = 


= 90° 


25' 21" 


/JL — 6^° 8' 18" 


' V = 


-- 152° 


8' 8". 


2. F 


X 


y 


z 


a 




P 


7 


3 


2 


— 1 


— 5 


10° 




100° 


* 


— 1 


— 3 


— 4 


7 


97° 




30° 




— 3 


5 


7 


6 


32° 






59° 


7 


6 


3 


2 






45° 


60° 


& 12 


— 14 


13 


2 


49° 




33° 




10 


3 


2 


1 


30° 






60° 


5 


4 


3 


2 


36° 






53° 


7 


2 


1 


— 3 


72° 






45° 



FORCES IN SPACE. 31 



4. 



5. 



F 


X 


y 


z 


a 




P 1 


% 


3 


1 


2 


3 


37° 




75°' 




5 


2 


— 3 


5 


87° 




' 23° 




-7 


— 2 


— 1 


3 


40° 


o 




61° 


9 


— 1 


4 


2 


44° 


7' 




47° 


- 6 


3 


3 


— 1 


90° 






90° 


3 


2 


— 4 


— 6 


30° 


5' 


79° 55' 




- 5 


— 3 


— 1 


— 4 


25° 


37' 


80° 




8 


— 2 


3 


— 7 


61° 


5' 


30° 




7 


11 


12 


10 


10° 




98 p 


14° 


9 


13 


21 


8 






77° 


14° 


16 


4. 


7 


9 


160° 




73° 




14 


6 


3 


2 






158° 


81°. 


23 


9 


8 


7 


■ 47° 




58° 




48 


1 


X 

4 


1 


32° 


1' 




81° 9' 


17 


— 4 


3 


— 2 


5o* 






55° 


65 


5 


i 

13 


eV 


63° 




63° 




14 


1 

<2 


JL 

7 


i 

14 


72° 




72^ 





Suppose after we have reduced one set of forces to a single 
resultant force acting at O, and a couple ; OP = R, to be the re- 
sultant force, and GC = M, the axis of the resultant couple; and 
let d be the angle between these lines. 

By a principle of projections we know that the projection of the 
straight line joining two given points on another line is the same 
as the projection of any broken line joining the same two points. 
Now let us project OP = R on the line OC, we obtain OD = 
R cos 0. Then project in its stead the broken line OABP, and 
the length of this latter projection must be equal to the former. 
Now OA, OB and OC, are the coordinates of P, and make with 
OC the same angles as the axes OX, OY and OZ, that is /, ,a and 
v, respectively ; hence the length of the projection is equal to 
OA cos A + OB cos ,u + OC cos v, 
but OA = R cos a r ; OB = R cos (3 r ; OC = R cos y x 



32 CONDITIONS OF EQUILIBRIUM. 

.\ R cos = R cos a Y cos A + R cos (3f cos p + R cos j> r cos v. 
.\ cos 6 = cos « r cos / + cos £> r cos (x -\- cos y x cos v. 

I. When cos = 0, or = 90°, the force lies in the plane of 
the couple, and we can then reduce them to a single force, acting 

M 

at a distance from O equal to fr> and parallel to R. 

II. If X = a T , ;i = f> r , and v = y T , that is, if the axis of the 
couple is parallel to the force, then we have a force of translation, 
and a couple causing rotation around the direction of the force as 
an axis, and we can simplify it no farther. 

III. If the axis of the couple is oblique to the direction of the 
force we can reduce them to a single force, and a couple acting as 
in the former case. For the couple M may be resolved into two 
components, one of which acts in a plane perpendicular to the 
direction of R (M cos #), and another acting in a plane containing 
the direction of R; and the latter on being combined with R gives 
an equal and parallel force at a distance from the line of action of 

M sin 

the 1st = — p 

IV. When R = there is only a couple. 
V. When M = there is a single force. 

CONDITIONS OF EQUILIBRIUM. 

The conditions of equilibrium are evidently that there shall be 
no motion of translation and none of rotation, that is, 
R = 0, and M = 0. 

... R2 = R x 2 -I- R 2 2 _|_ J> 3 2 = (^ 

and M 2 == M/ 2 + M 2 2 v + M 3 2 = 0. 
... R 1 = 0, R 2 = 0, R, 8 = ; M*= 0, M 2 = 0, Mo = ; 
and these are the six conditions of equilibrium. 

EXAMPLES. 

In all the examples of the preceding article reduce the result to 
a single force and a couple, as in III. 



PARALLEL PROJECTIONS. 33 

PARALLEL PROJECTIONS. 

In order that one figure may be a parallel projection of another, 
two conditions must be fulfilled. 

First. Each point of the one must have a corresponding point 
in the other ; that is, there must be some law connecting the two 
figures, such that as soon as a point is given in the first a point 
is determined in the second corresponding to it. 

Second. For each pair of equal and parallel lines in the one 
there must be a corresponding pair of equal and parallel lines in 
the other. 

Thus, suppose we draw in Fig. 23 any two parallel lines, and 
take AB and CD equal. Xow determine the point, A', B', C, D', 
of (Fig. 2F) respectively corresponding to A, B, C, and D, and if 
the second figure is a parallel projection of the first, A'B' is equal 
and parallel to CD'. 

This is the geometric definition, and I now proceed to show that 

x ' y z ' ' 

if the algebraic definition, — = «, — = 0, — = c, is fulfilled, 

° x ' y z 

then the geometric is also. I shall limit myself to a plane, and it 
will be easy for the student to extend it to space. Suppose AB 
and CD to be a pair of equal and parallel lines in the first figure 
and that A', B 7 , C, D' are the points corresponding to A, B, C and D 

respectively, and suppose the algebraic condition, ^ = #,^7 = 

to be fulfilled J Coordinates being as follows : — of A (x, y), of B 
(,/. /). of C {x'\y"), of D (#"', f% of A 7 (X, Y) &c.\, then the tri- 
angles AEB and AFD can easily be proved equal to each other ; 
hence AE = CF, but AE = x* — x, and CF = x" — x" ; again 
from the conditions we have 

x x r x^ x m 

X ~ — x / — ^7/ — ^777 — a 

, and since x — x = x!" — x" 



therefore X' — X = X'" — X", or A y E' = CF\ so also B'E' : 
D'F'. and hence the triangles A'E'B' and C'F'D' are equal, and 



M PARALLEL PROJECTIONS. 

A'B' is equal and parallel to CD', which proves the geometric defi- 
nition which may be otherwise stated thus : — 

The parallel projection of two equal and parallel straight lines is a 
pair of parallel and equal straight lines. 

This last definition renders the demonstration of the Proposi- 
tions enunciated by Eankine very easy. I proceed to give the 
demonstrations. 

I. If two equal and parallel straight lines have their parallel 
projections equal and parallel, it follows that equal portions of the 
same straight line project into equal portions, and hence that the 
parallel projections of two given portions of a straight line are in 
the same ratio as the portions themselves. 

A particular case of parallel projections is the case of project- 
ing a figure on any plane by parallel projecting lines, as in Fig. 25. 
CD is a parallel projection of AB, and in this case one-half of CD 
orone-third of CD would be the parallel projection of one-half of AB 
or one-third of AB, &c. Hence in this case also the proposition is 
clear. 

II. The same reasoning will apply to this : — 

If equal portions of parallel lines project into equal parallel 
lines, the parallel projection of a system of parallel lines must be 
a system of parallel lines whose lengths bear to each other the 
same ratio as the lines themselves. Thus, if AB and CD are 
equal and parallel it is easy to prove that A'B' and CD' are also 
equal and parallel ; or if AB and CD are parallel, but not equal, 
that A'B' and CD 7 are also parallel and in the same ratio as 
AB : CD. 

III. Find the parallel projection of each of its vertices, 
and join them, and we have a closed polygon. 

IV. The parallel projections of two opposite sides will be equal 
and parallel, and hence the resulting parallel projection will be a 
parallelogram. 

V. The parallel projection of each of its faces will be a paral- 
lelogram, and hence the resulting parallel projection will be a par- 
allelopiped. 



PARALLEL PROJECTIONS. 35 

VI. Divide up the plane surfaces into an indefinite number o{ 
small rectangles, all equal to each other ; the parallel projections of 
these rectangles wil5 be parallelograms equal to each other, and 
the number of rectangles in either of the first surfaces is equal to 
the number of parallelograms in its parallel projection. 

VII. Divide the surfaces into small rectangular parallelopipeds 
and the same reasoning holds. 

Demonstration of 63 and 64 of Rankine. 

Suppose we have a pair of forces (Fig. 26), P and Q, applied at 
the points A and B of a body, and that the lines P' and Q', and 
the points A' and B* are parallel projections of P, Q/A and B. 
Xow if C is the point of application of the resultant where 
AC : BC : : Q : P, the parallel projection of C will be C, where 
A'C : B'C : : Q' : P' : : Q : P, by proposition I. of parallel projec- 
tions ; hence the resultant K/ of P / and Q 7 will be the parallel pro- 
jection of K the resultant of P and Q. 

From this both 63 and 64 follow. 

Demonstration of 65 and 6Q follow in like manner from 
IV. and V. 

EXAMPLES. 

Given (Fig. 27) AC = 5, AB — 7, and the length of the par- 
allel projection of AB = 10, find those of AC and BC. 

Given (Fig. 28) AB — 5, A'B' = 4, A"B" = 3, A'"B /7/ = 2, 
and the parallel projection of AB = 7 find thosa of the others, 

3. Given the area of a circle of radius a equal to -a 2 , find that 
of an ellipse inscribed in the circle whose semi-minor axis is b} 

4. Given the volume of a sphere of radius a equal to f t: a?, 
find that of the ellipsoid whose semi-axes are a, b, c respectively. 

%f yf J) 

1 The ellipse is a parallel projection of the circle for — = 1, '— = - where (a/, y f ) 
is a point in the ellipse, and (x, y) in the circle. 



36 DISTRIBUTED FOCC'ES. 

DISTRIBUTED FORCES AND CENTRE OF GRAVITY. 

We have heretofore considered forces acting at a single point of 
a body, a condition which never exists in nature, as all forces act 
on some extended surface or volume ; but our previous results are 
necessary here, since all distributed forces can be reduced to a 
single resultant force acting at a definite point, a resultant couple? 
or a combination of the two. 

Experiment and observation teach us that all bodies celestial 
and terrestrial are attracted towards the centre of the earth. 

In machines and structures, on account of their small size in 
comparison with the earth, the force of gravity may be conceived 
to act on each particle of a body in parallel lines. 

Thus the weight of a body is the resultant of a set of parallel 
forces, and the centre of gravity is the centre of these parallel 
forces. If the weight per cubic foot of a homogeneous substance 
be known, and also the number of cubic feet in the body under 
consideration, the total weight is at once ascertained by multiply- 
ing these two numbers together. 

What Rankin e calls in these articles specific gravity is the 
weight of one cubic foot of a body. 

EXAMPLES. 

1. What is the weight of a cubical block of granite, whose 
edge is 6 f t ? 

Solution. Volume = (6) 8 = 216 cubic feet ,\ Weight is equal 
to 216 X 172 = 37152 pounds, taking the weight of granite per 
cubic foot to be 172 pounds. 

2. 1 What is the weight of a rectangular block of magnesian 
limestone, whose three edges are 4.7? 3.1, 5.6 feet respectively? 

3. What is the weight of a hollow cast iron column of the fol- 
lowing dimensions, height ~ 10 ft., outer diameter of base ^ 
1 ft. 2 inches, inner diameter = 10 inches ? 

1 A table of weights of different substances per cubic foot will be found in Kankine'si 
App. Mech. 



CEXTRE OF GRAVITY. 37 

4. What is the weight of a bar of silver whose length is 
10 inches, and whose cross section is a rectangle (edges are 3.G 
and 1.2 inches respectively) ? 

6. How deep a box will be required to hold a ton of anthra- 
cite coal, the base of the box being a rectangle (edges 3 and 6 ft, 
respectively) ? 

SPECIFIC GRAVITY. 

The Specific Gravity of a substance is the ratio the weight of 
any volume of that substance bears to the weight of an equal vol- 
ume of another substance taken as a standard. Thus a cubic foot 
of lead weighs 712 pounds, while a cubic foot of dry sand weighs 
about 89 pounds, this shows that lead contains 8 times as much 
matter as sand in the same space, or it is 8 times as dense, and if 
sand were taken as our standard of specific gravity that of lead 
would be 8. 

The substance usually taken as our standard of specific gravity 
is distilled water, at the temperature of maximum density, a cubic 
foot of which weighs 62.425 pounds; so that the specific gravity 
of lead is el-V^s == ilzk 

The weight of a cubic foot of a substance can be obtained by 
multiplying its specific gravity, referred to water as a standard, 
by 62.425. 

CENTRE OF GRAVITY. 

The centre of gravity of a body has already been defined as the 
centre of the set of parallel forces which make up the weight of 
the body ; or in Rankine's words, the point always traversed by 
the resultant of the weight of the body. 

Rankine's Art* 71 and 72, should now be read* 

GENERAL FORMULA 

Suppose we have a connected system of bodies whose weights 
are W x , W 2 , W 3 , etc.. respectively, and that when referred to a set 
of rectangular axes the coordinates of their individual centres of 



o8 CENTRE OF GRAVITY. 

gravity are fa , y x , %), (x 2 , # 2 , z 2 ), (;r 3 , y 3 , z s ), etc., respectively, 
it is the same as though we had a set of parallel forces equal to 
W, , W 2 , W 3 , etc., the coordinates of whose points of application 
are (x x , y t •, 2j), (# 2 , #2? 2 2 ), etc ^ respectively, and the centre of 
these parallel forces is the centre of the system. 

To determine the coordinates of the centre of these parallel 
forces, we have the resultant moment equal to 

(W, + W 2 + W 3 + etc.) x = W& + W 2 ^ 2 + W3Z3 + etc., 

or x 2 W =± 1 Wx. 

Where (#, y, z) are the coordinates of the centre ; so also y 2W 

= I Wy, and z 2 W = 2 Wz. Hence, for the coordinates of a 

system of bodies we have the following general formula : 

-__2Wx -_ 2Wy -__2Wz ' 

^W ^ _ Tw' z ~~^w' 

EXAMPLES. 

1. Suppose a rectangular homogeneous plate of brass (Fig. 29), 
where AD =12 inches, AB = 5 inches, and whose weight is 
2 pounds, to have weights attached at the points A, B, C and D, 
respectively, equal to 8, 6, 5 and 3 pounds, find the centre of grav- 
ity of the system. 

Solution. Assume the origin of coordinates at the centre of the 
rectangle, and we have 



W a = 2, W 2 = 8, W 3 = 6, W 4 = 5, 


w 6 = 


3. 


x x =0, x 2 — 6, x 3 — 6, # 4 = — 6, 


x 5 = 


— 6. 


yi = °. 2/2 = h ys = — h Pi — — h 


y s — 


#• 


.-. 2 Wx — + 48 + 36 — 30 — 18 = 


: 36. 




y Wy — + 20 — 15 — 121 -f 7^ =± 


0. 




^W = 2 + 8 + 6 + 5 + 3 = 24. 






.. x — 24 — ± 2 , y — 24 — V ' 







Hence the centre of gravity is situated at a point E on the line 
Ox, where OE = 1£. 

2. Given a uniform circular plate of radius 8, and weight 12 
pounds (Fig. 30). 



, 



SLENDER RODS. 39 

At the points A, B, C and D weights are attached equal to 10, 
15, 25 and 23 pounds respectively ; also given AB = 45°, BC = 
105°, CD = 120°, find the centre of gravity of the system. 

HOMOGENEOUS BODIES. 

In the case of a single body it is the same as though we had a 
system of bodies, the individuals of which are its separate parti- 
cles, and the general formulae still apply when W 1? W 2 , &c, repre- 
sent the respective weights of the individual particles. If u\. w 2 , 
&c, represent their weights per unit of volume, and V 19 V 2 , &c, 
represent their volumes respectively, W x == u\ V l5 W 2 = u\ Vc, 
W 3 == w 3 V 3 , &c, and when the body is homogeneous, that is, of 
the same density throughout, w 1 = w 2 w B , &c, and the formulae be- 
come 

- I tvYx 2 ivYy - 2 wYz 

x= 2wV' y= 2wY' Z= 2wV 

or, since w is a common factor in numerator and denominator, 

-_:Yx -_ IYy_ - _ 2 Yz 

x — £ V ' y ~ 2Y' z ~~ 2 Y ' 

where 2 Y is the volume of the body. 

CENTRE OF GRAVITY OF SLENDER RODS. 

Suppose we have a uniform rod, so slender that all its particles 
may be considered, without sensible error, to lie in the same plane. 

Let a be the sectional area, then the length of any small portion 

being A s its volume will be approximately a J s, and substituting 

this for V in the formulae (2), we have 

- 2 ax J s I x J s - 2 y A s 

x = -T- ; — = x . ,. — » y = v , — * 

2. a J s 2 *j s J 2 A s 

Now the smaller we make the parts the nearer are these for- 
mulae correct, and hence we have the true formulas, 

fxds - fyds 
x = r ■ ' y = r , 
fas J fas 

Bear in mind that ds = *J \dx 2 + dy' 2 ) = y (l + -4-^\dx, 



40 CENTRE OF GRAVITY OF A PLATE. 

EXAMPLES. 

1. Find the centre of gravity of a quadrant arc of a circle 
radius r. 

Here we have x 2 + y 2 = r % hence xdx + ydy = ; or 

<fe y X <to 2 ^ f y 1 y 1 

t xdx 

~dx = j (f _ x * ) ; hence we have 

p y rxdx . r r rdx J y/ 2X ) r 



1 r sin - r 



2r 



2 



and from the symmetry of the figure y 

2. Find the centre of gravity of a straight line of length I by 
the above formula. 

3. Find the centre of gravity of a semiellipse, both as cut off 
by the minor, and second by the major axis. Semiaxes are a and b. 

4. Find the centre of gravity of an arc of an elliptic quadrant. 

CENTRE OF GRAVITY OF A PLATE. 

The centre of gravity lies of course in the central plane of the 
plate. Divide it into small rectangles, all equal, by lines parallel 
to the axes of x and y, whose distances apart are A x and A y, 
respectively, and suppose the thickness to be Z, we have for the 
elementary volume / A x A y ; and substituting this for V we have 

- IxlAxAy ZxAxAy — XyAxAy 

x = HAxAy ' 2 Ax/ly 1 y ~ 2 A x A y ' 

which are the approximate formulas from which, by making A x 
and Ay smaller and smaller, we derive the correct ones, 

ffxdxdy — ffydxdy 

ffdxdy y ffdxdy 



PARALLEL PROJECTIONS. 41 



By performing one integration with reference to y they become 



Jid 



fxydx - J 2 UUy # _ fxydx _ fy 2 dx 

fydx ' y ~~ "/yd* , 01 x — ^^ > 3/ — 2 y yrfa . > 



and if we put in the limits y 1 and y 2 for y, we have Rankine's 
formulae, viz., 

-_ /*(& — yi)<fc - = f{y2—y*)dx 
x fin* — yi)d* y 2S(3/2 — yi)dx 



EXAMPLES. 

1. Find the centre of gravity of the portion of the parabola 
y 2 = 4 X between the two ordinates, where x = 4 and x = 9, and 
above the axis of a?. 

Solution. 

/»9 3 



' 2fJ dx " (fa^~" V_ 4 f 
35 2 5 

£ — 3 • 2_1_1 fi fifiQ _1_ 

— 533 2 3 — 5 "1 9 " — 0.000 -j- 

Note on Rankine's App. Mech., Art. 81, second Approx. 

Area of trapezoid ABCD (Fig. 31) = (— -^ \2Ax = 

(u' + u"')Ax< 

Area of segment CGD = § of parallelogram CDEF 

- fABGH = f(tf"— U '\ U '" \dx. 
Hence the total area of two bands is equal to 

(V + u'")Ax + %u" Ax— %(vf + u'") Ax, or 
l(tf + W + tS") J x. 

Now let us calculate by this method the previous example, in- 
stead of integrating. 



42 



PARALLEL PROJECTIONS. 



u — & J 








u = x 2 


^a =4 
X x *±2 






= 8 
= 9.546 


m =2.121 


x( r = 5 




i 


11.095 


2.236 


x-r= h 




i 


12.898 


2.345 


" 6 




i. 


14.697 


2.449 


« 6£ 




u 


16.567 


2.549 


a 7 




a 


18.520' 


2.645 


a 7 1 

' 2 




a 


• 20.53<> 


2.738 


" 8 




i. 


22.627 


2.828 


" 8J 




a 


24.781 


2.915 


^ b =9 




u h 


= 27.000 


u h = 3.000 


^a + ^b 

2 2^ (plain) 
4 2' ^ (dotted) 

\Ax = 


h 


85.000 
133.878 
337.324 

6)506.202 


u a + w b = 5.000 

2 2u Y (plain) = 20.316 

4 2^ (dotted) == 50.672 

6)75.988 


r& 3 


dx 


r=r 


84.367 

. ~Z — 8 4"3«7 
. . Jj 1 2.6 6 4 


f x^dx = 12.664 
±= 6.661 + 



Now compute the same by the first method of approximation 
and we have 

u, + ^b 



17.5 2 u { = 151.27 



fa^rfaj = J (17.5 + 151.27) = 84.385 and 



/: 



2 dx = 



12.663 



since in this case 



u R + n h 



2.5 and 2 u = 22.826 



■•■ * = fl:fM = 6.664 + 

Note on Art 77 of Rankine. 

First, to prove that G G 3 (Fig. 32) is parallel to GjG 2 ; sup- 
pose both triangles removed, and that G is the centre of gravity of 
the figure ABDFE; then annex FEC, and the centre of gravity 
moves in the line GGj, to a point G such that GG X : GhGq = W 1 . 
W — W x . Now, instead of that, suppose the triangle I1DF to be 



PARALLEL PROJECTIONS. 



43 



annexed, and the centre of gravity moves instead to a point G 3 in 
the line GG 2 such that GG 3 : G 3 G 2 = W x : W — W P 

.*. GG : GiG = GG 3 : G 3 G 2 
hence the line G G 3 is parallel to G X G 2 . 

Secondly, from similar triangles we deduce G G 3 : GiG 2 = GG 
; GG X = W x : W , or 

W 

G G 3 = G X G 2 ^ 



SOLUTIONS OF RANKINES EXAMPLES OX CENTRE OF 

GRAVITY. 

I and II are sufficiently clear in the text. 

III. Take OD for axis of x, and take the moments about AB 
(Fig. 33) ; the elementary area will be 
FK (zJ x sin DOH) nearly, hence we have 

/OD 
FK (dx sin DOH) = 

/OD 
FK (x sin DOH) (dx sin DOH) 

or dividing out by sin 2 DOH we have 

/OD />OD 

FKdx = I FK x dx. 

Now FG : CG = AI : CI, or FG : OD — x = J (B 

(QD _ tT ) (B — b) ' 



6):OD 



FG = 



FK = 2FG + b = 



B-OD 



2 OD 
(B — 5) q; 



/OD 
FK a; ace 

t 

Ba? 



OD 



= B — 



B 



b 
OD - * 



/> - ^T) - 



FK cfe 



( Ba* 



(B -fl)g 
30D 



OD 



2 0D 



OD 

FKffe 



B 



(B + b) OD 



B + 6 



_ OD / B-bX 

— 2 \ 5 B + b) 



44 DISTRIBUTED FORCES. 

The area we know from Geometry is £ (B + b) DH 

= i (B + b) OD sin" ODE 
IV. Areas of similar triangles are to each other as the squares 
of their homologous sides .\ Wj : W 2 = xf : x* 

Wj x{_ 

_ W, § x l - -W 2 f^_ 2 W^ 1- ** x? ' 



w,— w, -*rw; ^^ 



-^ - 



W nr 2 

9 *^2 



2. _ - 



Y. Take moments about OY, then sin XOY divides out, and 
we have 

// xdydx I I ydydx 
_o_J_o_ . , J o J o 

I | <%£& | i dydx 

J t/ J J 

Let the equation of the parabola be y 2 = 4ax, and the formulae 
become 

/*i /» x i i_ a. /» x i 3 
ccyacc I 2a 2 x 2 dx * I o^tfe 2 f 
-J = J-* = «L_Q _ 5^1 3 

I yaa; I 2a 2 x 2 dx f x 2 dx § 

Jo Jo J 






/Xl /»X 

_ J 



a 2 ^ 2 



3/y2™ 2 — . a? 



ifO /»*i 1 /ill 1 3 4 8i/1 

2 1 ycfo 4a 2 / x 2 dx %x Y 2 

Jo Jo 

/y /»xi /»x! 

/ aycfo sin XOY = l ycfe sin XOY = 
o Jo Jo 1 

f a^ sin XOY .-. W == %wx l y 1 sin XOY. 

VI. 

/yi /»xi /»xj 

I xdydx I ^(i/j — y)cfc 

J J _ J_o __ 

^0 /»yi /»x x /.xi 

J I ^ %r& / (2/x — y)dx 

a*fe — xydx y^ — i x iVi 

_0 J = * __ 3 g, # 

yi f l dx — J ycfe ^ lXl 32/i^i 

Jo Jo 



CENTRE OF &RAVITY. 45 



c is substituted from its value above. \ 

r*Yi /» x i /» x i 



/yi /» x i /» x i 

y J o t/ o 

f f 1 dydx 2 f * (^ — y)<fc 

J y J o Jo 



y *J K dx '~ J* l4axdx ' _ yte - W _ y,^(i-4 ) = , v 

2y, f 1 db - 2 f 1 y* 2x lVl - 1^1 ^ 2 ~ *> ■ 

Jo i/O " 

/Xl 
ytfe sin XOY = ^i^/i sin XOY. 



VII. 

^rcos 9 



/r cos /»? 

^ 2 tan #& + / ccx/ (r 2 — a? 2 )aa5 

_o Jrcose v - y _. 

^° yy/x " ir 2 # 

r 3 cos 2 d sin r s sin 3 

3~~ 3 sin 
I^T0 = f ^-^— ; y = °> smce the centre 

of gravity is on the axis of symmetry. 

The area we know from Geometry is ^ arc X radius = \r(2r 0) 
= r 2 0. 

VIII, 

= fxydx = f T cos , ^ ^ " **) dx _ {~ *(** - x ^}\ cos , 
^° /2/^k " f / ^2 ___ ^ ^ Area of segment. 



f y (r 2 — ^ 2 ) 

J r cos 



cfe 



1^3 gm 3 ^ 1^3 g' n 3 Q gm 3 ^ 



sector — triangle r 2 — r 2 sin cos # 3 — sin cos 

~2~ 2 

2/o 2 /ycf* " 2- area ' r ^_ sin ^ os ^ 

r 3 cos 3 # 
~3~~ _ T ^ 2(1 -cos 0) -cos 0(1 -cos 2 0) 



§r 3 — r 3 cos + 



r 2 (0 — sin cos 0) — sin cos 

4sin 2 i — sin 2 cos 



3 — sin cos 



46 



CENTRE OF GRAVITY. 



Area = sector — triangle = 



r 2 r 2 sin cos 



... W = \wr\d — sin cos 0). 
IX. Area of spandril =± rectangle + triangle — sector === 
(r sin 0) (r — r cos 0) + \r 2 sin cos — ir 2 = 
2 sin — sin cos — 



2 

^r sin 



/r sin /»r 

.W-*) ) rto S e XdydX ._ 

J V(r 2 -x2) J rcofe 

f x\rsmd-V(r 2 -x 2 )\dx \—$m + ^-x^V 
J r cm (2 ) r 



cos 



^ sin - 



area 
r 3 cos 2 0sin0 2r 3 



2 



sin 3 



area 
gj r 3 g i H 3 # 



area 



sin 3 



2 (2 sin - sin cos 0-0) 



= *r 



*' 2 sin 



sin cos — 



/r sm /»r 

^(r«-x«) Jrcosfl^ V ' % 



2/0 ffdyda 



area 



/ 



rcos 9 



(r 2 sin 2 — r 2 + a3 2 )cfe -j r 2 sin 2 0-sc — rV]+ -5 



x° 

3 .) r cos 



= yr 



2 • area r 2 (2 sin — sin cos — 0) 

3 sin 2 — 2 — 3 sin 2 cos + 3 cos — cos 3 



2 sin — sin cos — 







3 sin 2 0—2 sin 2 cos — 4 sin 2 ^ 



2 sin — sin cos — 
X. The centre of gravity is on the middle line 

VV \X\ VV 2^2 ' 



.*• 2/o = °> x o = 



Wx— W 2 



W7 1 

w, 



w, 



w x 
' but w, = 



— 1 



SLENDER RODS. 47 

sin 6 sin r s — r, s . sin 

Area = (r 2 — r x 2 ) .-. W = w(f — r 2 ) 9. 
XL When the elliptic arc is given we know the corresponding 
circular arc, for the coordinates of its extremity, being x and y, the 
x of the circle and ellipse are the same, and the y of the circle = 

-j-y of the ellipse, let x', 3/, be the coordinates of the centre of 

gravity of the circular arc, etc. ; then x = x', and y = — y\ 

XIII. Since the triangular face is parallel to the x axis, y = y ± . 
Volume, we know from Geometry, = x^y^ .\ W = icx^y^? 



•^0 ~~ «vi /»xi „. r™J~. ~ 2_. - 3*^1 • 



/yi /»-m o /»xi 
/ x 2 dxdy y 1 j x 2 dx 
Jo __ J 

f Vl f xi »*&•% " ft/**** " x iVi 
J J 

XIV. Find the value of ^/ in the terms of x, thus, 

y : # : : x x — x : Xl /. y = J/i ^ = y^l — — ) 

r 4a X \j S Xl x * } Xl 



Volume = ^, base X alt. = -J- wz^y^ 

XV. Equation of circle is x 2 -j- y 2 =r 2 .-. ^ = */ ( r 2 x i\ 

f x V(r 2 -x*) dx J j ( r 2 — a; 2 ) t t * 

W — 2 W - /" x V (r 2 — a;' 2 ) <fo = f w 2 " «' 
XVI. 

Wj = | M>rV, W 2 = § wrV', ^ = jg r, a; 2 = |g / 

W 2 — r'V but z" — ?"W^ _ 7* 



4 4 _3rr 



48 



CENTRE OF GRAVITY OF A PLATE. 



_ WjXj — W 2 x 2 



w 2 . ! 






w 2 



3- r' 
16 



~7a V 



3rr r 4 



~ 16 r 8 — r' 8 

XVII. Area of a section at a distance, jc, from the vertex is 

x 2 _, x 2 

A jr 2 /. Elementary volume = A t~ 2 dx sin 

A /» h 



— j x 2 dx 

h 2 Jo 



Sh 
= —r W = J wA/a sin # 



XVIII. Volumes of similar cones are to each other as 



£' 2 



av 8 > 



W ^ wAft sin # — %w Ajp h' sm = % wAh I 1 — p- ) sin 

XIX. Conceive the shell divided into a number of shells of 
small thickness, and let the variable radius be />, then area of zone 
is 2rcp ' />(cos £> — cos a), and the distance of the centre of gravity 
from the centre of the sphere is 



cos a -f- cos (3 

P 9 "" •"• x 



t I Ypdp 7r / /> 3 (cos 2 p> — cos 2 a) dp 
I Ydp 2n j /> 2 (cos (3 - — cos a)dp 



(cos 2 (5 — cos 2 a) 



cos « .+ cos (3 



( r 6 r rt o\ 
— — k J (cos ft — COS a) 

W = w I 2tt/> 2 (cos 2 ft — cos 2 a) dp 



2r,w / 



- / 3 ) (cos /5 - cos a) 



XX. \ Imagine the shell divided into thin plates by planes 
perpendicular to OX. We shall thus have a set of additive sec- 
tors in the outer sphere, and a set of subtractive in the inner. 



PRACTICAL PROBLEMS. 49 

Area of sectors are 6(r 2 — x 2 ) and d(r r 2 — x 2 ) respectively, 
and distances of centre of gravity from OX are 

tV^-^andfV^-,^ 

f x(r 2 — x 2 )dx — d f x{r f2 — x 2 )dx 

J_o ./_o 

T (r 2 — x 2 )dx — Of (r' 2 — x 2 )dx 

X Q ( r 4 _ / 4) y_ r '^ 

— ^0(r 3 — r fS ) ~~ 8 T 3 — r f * 

f r ( r 2 _ x Tyi dx _ d C* (/2 _ x iydx 

% = t sin -; ^-P = 

6\ (f — x^dx — Q \ {r'^ — x^dx 

% sin # = — • -o r« • — ^ — • 

3 %0(r* — r' s ) 16 r 3 — r' s 

SOME PRACTICAL PROBLEMS. 

In the preceding pages have been considered, first, the composi- 
tion and resolution of forces, in whatever way applied to a body ; 
second, the conditions of equilibrium ; and third, the determination 
of the centre of gravity of any body whatever. We are thus en- 
abled to solve an immense number of practical problems relating to 
the stability of structures and machines, and the strength and 
dimensions of their different parts. 

The determination of the suitable dimensions and forms to resist 
any known state of stress in any one of the parts, will be treated 
of hereafter, but it is the purpose of this chapter to compute the 
suitable distribution of the parts and stresses in each of them, re- 
sulting from any given condition of loading. 

There are two methods of proceeding in such practical prob- 
lems, first, by algebraic and numerical calculation, and second, by 
geometrical construction. Of these two the engineer finds now 



50 PRACTICAL PROBLEMS. 

i 

one and now the other the most convenient. The following prob 
Iem will exemplify them both : 

To determine the stresses in the rafters and in the tie beam, also 
the respective weights sustained by the struts at A and B in the 
frame ABC (Fig. 34), the weight W being applied at C, and the 
weight of the frame itself being disregarded. 

Solution. Assume the origin of coordinates at C, and the axis 
of x horizontal ; let t represent the stress (of compression) in .the 
rafter CB, and if that in CA, and H the tension in the horizontal 
tie beam, AB. Now the conditions of equilibrium require that 
2 P cos a = 0, 2 P sin a = and 2 Tp = 

The last is not necessary in this case, as the point of application 
of the force has been taken for origin ; the others give 
2 P cos a = t cos i — t r cos % f = 
2 P sin a — t sin i + t r sin i> — W = 

Whence we obtain 

W cos i , W cos i r 

t — - — 7T - j — jp- t r 






sin (i + i') ~ sin (i + i r ) 
These are the compressions in the rafters. 

mi -i * , . " ., "W" cos i sin i f 

The weight at A = t sm i == — ; — y. — -, — vt- 
fe sin (i -\- % ) 

W cos %'■ sin i 

u u B = t Sin l = : T7— — — — 

sin (i + * ) 

W cos i cos i' 

H = t cos v = t cos i = — : — T^—i — v^T 

sm (i -\- %) 

The geometrical solution is given in Fig. 35, where we draw the 
vertical line, EL, to any scale to represent the weight W. We 
now wish to determine the components of this weight in the direc- 
tion of the rafters, or, which is the same thing, the stresses in the 
rafters. To do this we draw through E and D the lines, EG and 
DG, respectively parallel to AB and CB, then we know by the 
principle of the Triangle of forces that EG and GD will repre- 
sent these required stresses (to the same scale as that to which ED 
was drawn). We next decompose EG into a horizontal and a ver- 
tical component, and we have GF for the stress in the tie bar, AB, 



PRACTICAL PROBLEMS. 51 

and EF for the portion of the weight acting at A ; so likewise 
FD is the portion acting at B, and by measuring these lines we 
have the magnitudes of the respective stresses, t = GD, t f — EG, 
H = GF. 

Wt. at A = EF ; Wt. at B = FD. 

Tf now the algebraic values of these quantities be determined 

from Fig. 35 we shall obtain the same values as before. Thus the 

sin GrDE cos i 

triangle GED eives EG = ED . 7 orx or t = W 



sin EGD sin (i + i) 



, ^ W cos V 

so t = GD = - — . . , .a , etc. 
sin \i-\-i ) 

EXAMPLES. 

1. Given W = 872 lbs., i = 35°, i r = 63°, to find the stresses 
in all the parts. 

Solution, 

872 cos 63° • w/i 

t = — . ■ = 399.76 

sm 98 

, 862 cos 35° 

f = . QQO = 721.31 

sm 98 

_ 872 cos 35° sin 63° 

Wt. at A = ; - 1 ^ = 327.47 

sin 98 

„ _ 872 cos 63° sin 35° 

Wt. at B = t—y^ = 642.70 

sin 98 

„ 872 cos 35° cos 63° ^ o ^ oK 

H = — ^o = 230.35 

sm 98 

The student should now make the geometrical construction, and 
thus compare the results obtained by each method. 

2. Given W = 953 lbs., »' = 43°, i = 47° 

3. Given W = 545 lbs., %' = 30°, i = 60° 

4. Given W = 763 lbs., i' = 27°, i = 59° 

Perform in each case the algebraic and also the geometrical so- 
lution, and compare the results. 

We will now take the case which more commonly occurs in 
practice, viz., that of a roof uniformly loaded. 



52 PARALLEL PROJECTIONS. 

Suppose (Fig. 36) AC and BC to represent a pair of rafters of 
a roof connected by a tie bar, AB, and we wish to determine the 
stresses in each of the parts. 

In this case we must observe that one-half the weight is borne 
by each rafter, and hence (W being the total weight) there is one 
force, ^W, acting at the middle of each rafter. Now of this ^W 
is directly supported at A, and ^W at B, while the remainder, 
£W + JW = £W, is the weight at C. 

In the case before us also i = i\ hence the results of the pre- 
ceding case become 

iW cos i _ W 



t = t r = 



sin 2 i 4 sin i 

H 



^W cos 2 i W 



sin 2 i " 4 tan i 
The weights sustained at A and B respectively are -~-. Tlie ge- 
ometrical solution is given in Fig. 37. In this case EGD becomes 
an isosceles triangle. 

EXAMPLES. 

1. Given W = 782 lbs., i = 42° find t, f, and H. 

782 
Solution. t = ~r—. — ™ = 292.17 

4 sin 42° 

H = rtan^F = 217 ' 12 

2. Given W — 1054 lbs. i = i r = 75° 

3. Given W = 1272 lbs. i = i' = 83° 

4. Given W = 725 lbs. i = i r = 45° 

5. Given W = 324 lbs. i = %' = 30° 

6. Given W = 572 lbs. i = i r = 60° 

Give both the algebraic and the geometrical solution in each 
case. 

As another example let us take a Warren Girder (Fig. 38) sup- 
ported at the points A and B, and loaded uniformly throughout 






PRACTICAL PROBLEMS. 53 

its whole length with a load equal to g lbs. per foot, long measure 

(including the weight of the girder itself). We wish to determine 

the tensions in the upper, the lower beam, and in the diagonal 

braces. The total load of the girder will be gl (7 being its length), 

gl 
hence the supporting forces at A and B are each equal to -y, and 

act upward. 

Suppose now we wish to find the stress R L along the upper bar 
at D ; that (R 2 ) along the lower bar at C, and the stress (R 3 ) in 
the diagonal brace DF. 

We must observe that the forces tending to turn the cross 

gl 

section, DC, are first the supporting force, ^-, acting at A, and 

whose moment with respect to D is ~ • AC = -^ ; secondly the 

force gx uniformly distributed over the portion GD, and which 

may be assumed to act at E, the middle of GD ; hence the mo- 

ax 2 

-, and hence the total bending mo- 



ment of this about D is 


— -~— , and 


. glx gx 2 
ment is -~- — -5- or 






f Q-* ) 



Now, decomposing the tensions parallel and perpendicular to 
GD, and imposing the conditions of equilibrium, we have 

R x + R 2 + R 3 cos i = and R 3 sin i = 5- — gx 
also taking moments about D 



v - 9X 



a — x), r 3 = -A-. ( T — x) 

v ' ° sin 1 \2 / 

R! = — R 2 — R :i cos i = — ^r (Z — x) — g cot i I 5 — x\ 
These give us the tensions at a joint situated at a distance x 



54 STRESS. 



from A. The tensions at the middle would be found by making 

I 

x — ^, and we have 

** ~ 8A' iX3 ~~ U ' ^ 8£' 



EXAMPLES. 

1. Given g = 700 lbs., I = 60 ft., h = 5 ft., < = 45°, find R l9 
R 2 and R 3 at the middle of the girder, and also at the top of the 
fourth brace. 

Solution. 

700 X (60) 2 
At the middle R 2 = — Rx = — 8 x v 5 J = 63000 lbs. R 3 = 

At the top of the fourth brace x = 20 ft. 

700X20X40 __„ 
•\ R 2 = -r-ry = 56000 lbs. 

R 3 = /0 ° * Q = 7000 V2 = 9899 lbs. 
6 sin 4o 

2. Given # = 600 lbs. I = 50 ft. h = 5 ft. i = 45° 

3. Given # = 930 lbs. I = 70 ft. A = 7 ft. i = 45° 

4. Given # == 900 lbs. I = 80 ft. h = 8 ft. £ = 60° 

5. Given g = 850 lbs. Z = 75 ft. h = 7 ft. t = 70° 
Find in each case the values of R x , R 2 and R 3 at the middle, 

and also at the top of the fourth brace from the middle. 

STRESS. 

Under the head of stress we have to consider forces distributed 
over surfaces, and not acting at a single point. 

Thus if we imagine an iron rod suspended from the ceiling with 
a weight of 1000 pounds attached to it, that weight exerts a stress 
on the particles nearest to it, those on their neighbors, and so on : 
and thus the whole rod is in a state of strain. 

The intensity of a stress is its amount per unit of area. 

Thus suppose the above rod to be divided into two parts by an 
ideal plane perpendicular to its axis, and that the area of this nor- 



STRESS, 09» 

mal cross section Is 12 inches ; then the whole stress tending to 
separate the lower from the upper part is 1000 pounds, and hence 
the intensity of the stress is L j%- = 83J pounds per square inch. 

EXAMPLES. 

1. What is the intensity of the stress per square inch on a nor- 
mal section of a rectangular rod (sides 10 and 6 inches), to which 
a weight of 1563 pounds is attached? 

2. What is the intensity of the stress on a normal section of 
the rectangular bar of a plough (sides of section 1.5 and 3 inches) 
when a dynamometer placed between the horse and the plough in- 
dicates a pull of 3000 pounds ? 

3. What is the total amount of the stress in a rod whose nor- 
mal cross section is a circle of radius 2 inches, when the intensity 
of the stress per square inch is 5.7 pounds ? 

If, on the other hand, the intensity of the stress is variable (dif- 
ferent at different points of the surface) we must divide the area 
over which it acts into small rectangles whose sides are A x and 
A y, parallel to Ox and Oy respectively, and if the intensity of the 
stress at the point (x, y) be represented by p (p being variable), 
the amount of stress on this rectangle will be p A x A y nearly? 
and hence the total amount of stress on the whole area will be 
P = // pdxdy ; also the area is S = ff dxdy, hence the mean 

P //pdxdy 
intensity is ^o^g- = ffdxdy ' 

UNIFORM AND UNIFORMLY VARYING STRESS. 

The two kinds of stress that we meet with in practice are the 
Uniform and the Uniformly varying. 

By a uniform stress we mean one whose intensity is the same at 

all points of the surface on which it acts. This is the case if we 

have a rod suspended from the ceiling with a weight attached to it; 

' if we imagine the rod cut through by a plane perpendicular to its 

i axis, the intensity of the stress at all points of this section is the 



56 UNIFORMLY VARYING STRESS. 

same. This is also the case in tie bars and braces ; columns and 
struts which support a weight, whose length does not surpass cer- 
tain limits. In a tie bar the stress is one of extension, and in a 
strut, of compression. 

To illustrate Uniformly Varying Stress suppose we have (Fig. 
39) a beam firmly imbedded in a wall at one extremity, and loaded 
at the other with a weight, W ; the effect of this weight is evi- 
dently to bend the beam, and in this bending the upper fibres are 
extended and the lower compressed ; there must be, then, a set of 
fibres or a lamina somewhere between the two that is neither ex- 
tended nor compressed. This lamina is called the Neutral Lam- 
ina. 

Suppose the beam to be cut by a pair of parallel planes at right 
angles to its length before bending ; it is evident that these planes 
will diverge after the bending. Let the planes be DE and HG. 
These include a set of fibres of the same length, but when the 
weight is applied and the beam is thus bent the cross section HG 
takes the position LM, such that all the fibres above FK are elon- 
gated, while those below are compressed (FK being a portion of 
the longitudinal section of the neutral lamina CP). 

Moreover the amount of tension on any fibre (either of exten- 
sion or compression) is proportional to its distance from the neu- 
tral lamina, as can be easily shown. 

Here, then, the stress on the cross section, HG, tending to sepa- 
rate the part of the beam beyond this cross section from that be- 
hind it, varies uniformly from the line in which this cross section 
cuts the neutral lamina or the line represented in section by K, 
and this is called the Neutral axis of the cross section HG. 

When the beam is not bent beyond the limits of elasticity the 
stress of extension and that of compression are equal, which 
shows that this neutral axis must pass through the centre of gravity 
of the cross section. Thus, if the cross section of the beam is a 
rectangle, AB CD (Fig. 40), the neutral axis will be EF, if (Fig. 
41) it is a circle, the neutral axis will be AB parsing through its 
centre, etc. 



GEOMETRICAL REPRESENTATION OF STRESS. 57 

As a second example let us take the case of the vertical wall of 
a reservoir. It is an established fact that the pressure of any 
body of water at any point is proportional to the depth of the 
point below the free upper level. Thus, suppose AB (Fig. 42) to 
be a section of such a wall, the pressure at the top (supposing the 
free upper level to be even with the top of the wall) is nothing ; 
suppose now that at the bottom to be represented by the line CB ; 
then that at any point D will be represented by ED such that 
ED : CB = AD : AB. Here, then, we have again a case of uniformly 
varying sfress varying uniformly from the free upper level of the 
water. 

So likewise in a revetement wall which sustains a bank of earth 
we have again a case of uniformly varying stress. This will give 
the student an idea of some of the cases in which he has to make 
use of the calculations of both uniform and uniformly varying 
stress. 

GEOMETRICAL REPRESENTATION OF STRESS. 

^\Ye have already seen that the simplest way to represent a force 
acting at a certain point of a body is by means of a straight line 
drawn from the point of application of the force in the direction of 
the force, and containing as many units of length as there are units 
of force in the given force. 

Now stress is a set of parallel forces distributed over the entire 
surface on which it acts. These forces, it is true, are balanced by 
another equal and opposite set of forces, which are brought into 
action by the cohesion of the body, but nevertheless since they are 
forces a straight line suggests itself to our mind as the proper sym- 
bol to represent each one of them. These lines must evidently be 
proportional to the amount of stress acting on the unit area, or 
proportional to the intensity of the stress ; thus at a point where 
the intensity is twice as great our line must be twice as loug. 

Now since the stress is distributed over the whole surface on 
which it acts we shall have an infinite number of such lines, or a, 
volume of a prismatic shape, whose lower base is the surface on 



58 GEOMETRICAL REPRESENTATION OF STRESS. 

which the stress acts, and whose upper base is formed by the ends 
of all these lines representing the stress at each point. 

In the case of Uniform Stress the upper base becomes parallel 
and equal to the lower, as in Fig. 43 ; if the intensity of the 
stress at auy point D of the circle ABC be represented by DE, 
the stress at any other point (supposing it to be uniform) will be 
represented by a line parallel and equal to DE, and we have thus 
a cylinder which is a suitable representative of the whole stress on 
ABC. 

If the stress be not uniform the upper surface will not be paral- 
lel to the lower, but in the case of uniformly varying stress it 
will be a plane inclined to the plane of the surface on which the 
stress acts, as in Fig. 44, where the truncated cylinder, ABCDEF, 
represents a stress acting on the plane of the circle ABC, and va- 
rying uniformly from the line 00'. 

Suppose that on the surface, ABCD (Fig. 45), we have a stress 
varying uniformly from the line GH ; we draw FC to represent 
its amount at C, and then the stress will be represented by the vol- 
ume ABCD — EF. Now if we draw through O, the centre of 
gravity of ABCD, a line perpendicular to the plane, and through 
P, where it meets the upper surface, pass a plane parallel to 
ABCD, we shall have a cylinder ABCD — IK equal in volume to 
the former figure for the wedge EIP = KFP. Now this cylinder 
will represent a uniform stress whose resultant acts of course 
through O ; and the two wedges in question represent a stress 
uniformly varying from the line LM, whose resultant is a couple 
acting in a plane perpendicular to LM. 

Combining these two resultants would merely have the eifect of 

shifting the former resultant to a point situated in the line AC 

M 
perpendicular to GH at a distance L = -p- from O. 

I have taken the stress normal to the surface to illustrate, but 
if it were oblique the line AC would not necessarily be perpendic- 
ular to GH. The line DB parallel to GH is the neutral axis, and 
AC, which passes through O, and the point of application of the 
resultant stress, is the conjugate axis. 



CENTRE OF STRESS. 59 

CENTRE OF STRESS. 

We must now proceed to find the coordinates of the point of 
application of the resultant stress, or in other words the centre of 
stress (the whole is supposed referred to a pair of rectangular 
axes). 

Let p denote the intensity of the stress at the point whose 
coordinates are x and y. Divide up the surface into small rectan- 
gles, whose sides are Ax and Ay respectively, and we have for the 
force acting on one of these rectangles pJxAy nearly, and for its 
moment with reference to the y axis pxAxJy nearly ; hence for the 
whole surface we have the entire moment in reference to Oy 
equal to x ffpdxdy = ffpxdxdy, and in reference to Ox 
y» ffpdxdy = ffpydxdy, (x , y ) being the coordinates of the 
centre of stress ; hence we have 

SSv xdxd y % ffpy dxd v, 

Xq ffpdxdy #> ffpdxdy 

which must be integrated between the proper limits. 

The centre of stress is the same as the projection on the plane 
in question of the centre of gravity of the ideal solid mentioned 
in Art. 88 of Rankine. 

EXAMPLES. 

1. Where is the centre of stress for a semicircle, the stress 
varying uniformly from the diameter ? 

Solution. Here the stress varies as its distance from y, or as 
the abscissa of the point, hence p = ax, and the equation of the 
circle is y 2 -\- x 2 = r 2 , or y = ^ V(?'" 2 — x 2 ). 



f V(r2 " ^ f ax 2 dydx 2a f xW(r 2 - x*)dx 

I I axdydx 2a J ;rV(V 2 — x 2 )dx 

— l^r* — x 2 Y + tf V(r 2 — x 2 )dx 



T 



3 ] ,-3 8 /l 



GO 



UNIFORMLY VARYING STRESS. 



I/O = 



/ axydydx 

/V(r 2 -a.'2) /»r 
I axdydx 



0. 



-v(r 2 - ^ 2 )t> 

2. Find the centre of stress for a rectangle, the stress varying 
uniformly from one side, that is p = ax. 
Solution, , 

^3 



#o — 



I I pxdydx I J ax l dydx ab I arax T 

/>b /*c ~b /»c " /»e ~g 

j / pdydx \ I axdydx ab I xdx _ 

J0J0 J J Jo 2 

I 1 axydydx -~- I xdx -, 

/ I axdydx ab I j #& 

3. Suppose p = m + wx 2 is the law of the stress 

l / (ma; + nx s )dydx b i (mx + nx s )dx 

1 / (m -\- nx)dydx b j (m-f nx)dx 



fa 



yo 



mc' nc* 
~9~ 1 ^T 



2mc -}- nc B 2m + nc- 






2m -\- nc ~ ~ 2m + n° 



, and 



yo = 



/b /»c 
J (m + nx 2 )ydydx 

/»b />c 

I 1 (wz + nx*)dydx 



MOMENT OF UNIFORMLY VARYING STRESS. 

We have seen (Fig. 45) that a uniformly varying stress can al- 
ways be reduced to two, viz., a uniform stress whose intensity is 
the mean intensity of the entire stress, and 2d, a uniformly varying 
stress, varying uniformly from a line passing through the centre of 
gravity of the surface on which the stress acts. This line is called 
the Neutral Axis of the body for that condition of stress. This 



UNIFORMLY VARYING STRESS. 61 

second portion of the stress is equivalent to a couple, and its 
tendency is to make the surface on which it acts rotate around the 
neutral axis ; it is therefore necessary to know the moment of this 
couple, so that in structures and machines we may provide sufficient 
strength to resist it. 

Now if we call the intensity of the stress p, this means that at 
any point {x, y) the amount of stress (say the number of pounds) 
on one unit of surface would be p. But we cannot take a unit of 
surface at that point, for part of it would be at other points, hence 
take a small rectangular area AxAy, and we shall be always nearer 
the truth. Now if we took \ a unit of surface, the whole amount 
of stress would be \p ; so the whole amount of stress on AxAy is 
pAxAy, and if p = ax it is JP = ax Ax Ay. Now this is the force 
acting on this small rectangle, and we must proceed to get its mo- 
ment around each of the axes OX, OY and OZ (OX and OY lie 
in the surface). Now if these parallel forces which make up the 
stress are inclined to OX, OY and OZ, respectively, at angles «, j3 
and 7, the moments will be as follows (since z = o) (Rankine, 
Art. 91) : 

round OX z/P y ccfe y = axy cos y Ax Ay. 

round OY — A Px cos y = — ax 2 cos y Ax Ay. 

round OZ JP(x cos (3 — y cos a) = {ax 1 cos p> - axy cos a)AxAy. 

Now then the total moments round each of these axes will be 
found by integration, or we have as follows 
M x = a cos yffxydxdy =a cos yK. 

M 2 = — a cos yffx 2 dxdy = — a cos yl. 

M 3 = a (cos ftffx 2 dxdy - cos affxydxdy) = a (I cos £> - K cos «). 

And we then proceed to compound these couples as Rankine 
does in the Text. We thus know the force which has to be re- 
sisted by the strength of the material, or in a beam in equilibrium 
we know the resistance it opposes to any definite strain. 

For the moment of bending and twisting stress, see Text. 

These formulae all depend on the quantities I and K. The 1st 
has received the special name of Moment of Inertia, for reasons to 
be given hereafter. 



62 



MOMENT OF TNERTIA. 






The coordinates of the point of application of the resultant of 
the stress are, as we have already found, and as given in Rankine, 
ffxpdxdi/ af fx 2 dxdy ol 



#o 



ffpdxdy 
ffypdxdy 
ffpdxdy 



affxdxdy 
affxdxdy _ aK 
affxdxdy ' P ' 



perpen. to neut. axis. 



along ventral axis. 



MOMENT OF INERTIA. 

The quantity I = ffx 2 dxdy, called the moment of inertia of a 
plane surface about the axis of Y, is the sum of the products of 
the elementary areas by their distances from the fixed axis. 

The quantity J = ffy 2 dxdy is the moment of inertia of a body 
about the axis of x. 

Thus to get the moment of inertia of the plane area we divide 
it into small rectangles by lines drawn parallel to the atfis of x and 
y, and now if we wish the moment of inertia relatively to Y, we 
multiply each of these rectangles zlxzly by the square of its dis- 
tance from Y, and then integrate, and thus obtain ffx 2 dxdy; if 
we wish that about x we have ffypdxdy; if now we wish that 
about Z we multiply AxAy by the square of its distance from Z, 
or by OP 2 = r 2 = x 2 + ?/ 2 , and we thus have ff(x 2 + y 2 )dxdg, 
but this = ff x 2 dxdy + ffy 2 dxdy = I -J- J. This is what Ran- 
kine calls the polar Moment of Inertia. 



EXAMPLES. 

1. Find the moment of inertia of a circle about the diameter 
Y, — Y. 
Solution, 

M = f^ {r ^ X2 \ f xMydx = 2 f V(r 2 — x 2 )x 2 dx = 
~r 4 _ Try/ 4 



UNIFORM STKESS. 63 

Observe that the last integration need not be carried out, be- 
cause j ydx or / V(r 2 — x 2 )dx is the area of a semicircle, and 

-r 2 
this we know is equal to — ~- ' 

2. Find the moment of inertia of a rectangle about the axis 
Y, — Y passing through its centre of gravity, and parallel to one 
side. 

Solution. 

M = / J h * 2d ^= b 4 , xdx= ns + t) = T2 

2 2 2 

3. Find the moment of inertia of the figure (4G b). 
Solution. 

; M = 2 fX *»* "= 2b f, ** = ?(f - 1) « 

2 2 2 

This might also be found by subtracting the moment of inertia of 
the rectangle bll from that of bit ; thus M = ;~nr — To" == 

i iV 8 - * 3 > 

Note. Rend now Rankine, page 81, "from " In finding the moment of inertia of a 

surface of complex figure." 

I 

UNIFORM STRESS. 

| V 

In a rod or bar subject to a stress, either of extension or com- 
pression, by a load acting in the direction of its length, we have 
already seen how to estimate the intensity of the stress. 

The amount of stress which a rod of one square inch cross sec- 
tion will bear with safety is first determined by experiment. This 
quantity, which we will call K, varies for different materials. For 
wood it is about 1,000 lbs., for wrought iron 10,000, and for cast 
iron 3,000, when the stress is one of extension. 



C4 BEAMS AND GIRDERS. 

If then the cross section of any rod be represented by a, the 
weight it will bear will be W = AK, for it is just as though we 
had a bundle of rods, each of one square inch cross section, and a 
in number. 

EXAMPLES. 

1. How much weight will a wrought iron tie bar, the area of 
whose section is ^ square inch, bear with safety? Ans. W = AK 
= \ • 10,000 = 5,000 lbs. 

2. A wrought iron tie bar of circular cross section is to resist a 
stress of 15,000 lbs, in the direction of its length; what must be 
the radius of the cross section ? 

3. What must be the area of the cross section of the tie bar in 
each of the examples on triangular frames given in Figs. 34, 
36 and 38 : 1st, when made of wood, 2d, when wrought iron. 

APPLICATION OF UNIFORMLY VARYING STRESS TO 
BEAMS AND GIRDERS. 

A beam is imbedded at one end, and loaded at the other (Fig. 
39) ; it is required to find what load W it will bear with safety. 

Solution. The force tending to break the beam is the load W, 
and if the rupture were to occur at the cross section HG, the mo- 
ment of the breaking couple would be Wx (x being the perpendic- 
ular from K on the direction of W). The greatest value of x is 
evidently I ( = CP), hence the greatest moment of the couple 
tending to produce rupture is W/. 

This is resisted by the stress in the beam, which, as we have 
already seen, is equivalent to a couple whose moment is al (Ran- 
kine, Art. 92), and hence we must have WZ = al to provide 
against rupture. 

To determine the value of a, let S be the tension at the fibre 
situated farthest from the neutral line, and d the distance of this 
fibre from the neutral line, then 



BEAMS AND GIRDERS. 65 

The moment of inertia I, and the quantity d, depend only on 
the form of the cross section, and this quotient -r is called the 

moment of resistance of any cross section. The value of S is dif- 
ferent for different materials, and has to be ascertained by experi- 
ment, and it is called the modulus of rupture. It is not, however, 
safe to load a beam to its utmost, and hence in practice we sub- 
stitute for S only a portion of its value, which is determined by 
practice, and this portion is called the coefficient of safety (see 
Rankine, Art. 247). If we denote this coefficient by K we shall 
have the formula 

EXAMPLE. 

/ 

1. Suppose the cross section of the beam to be a rectangle, where 
b = 3 inches, h = 6 inches, I = 3 feet, we shall have 

K !_ *L t 1 ^ 8 _ , KM 2 

w — t" d — i i — *-"v ' 

In this case, W = JK %*f J = ^K. 

If the material is oak, for which the modulus of rupture is 
10,000, and for which we take 10-fold security, we have 

W == 500 lbs. 

If now, instead of having a load at the end, we have a uniformly 

distributed load, we may imagine it acting at the middle of the 

I 
beam, and then the moment of the breaking couple becomes W-^, 

and we have 

or, in words, the beam will bear twice as large a load when it is 
uniformly distributed over its length, as when applied at the end. 

K 

Tims in the above case the beam would bear W — § o7r ' 3(6) — 

1,000 lbs. uniformly, distributed over its entire length. 

9 



60 BEAMS AND GIRDERS. 

Observe that W should include both the weight of the beam 
and~the load. 

2. Suppose we have a beam supported at both ends (Fig. 47), 
and loaded in the middle with a weight W 7 the load supported at 
each of the supports is ^W. Rupture will evidently occur at the 
point where the load is applied, or the middle point, and hence we 
may regard each half of the beam as a beam firmly imbedded at 
one extremity and loaded at the other, so that the moment or rup- 
ture will be in this case, 

W I Wl TV7 ■ KI m 4K1 

T ' 2 = T •"• X = al = ~d> or W = ~ld ' 
If the load is uniformly distributed over the beam we may re- 
gard JW in each of the halves as acting at the middle, hence the 
moment of rupture is 

W I Wl _ 8KI 

2 ' 4 ~~ 8 v ™ — u m 

Thus a beam supported at both ends and loaded uniformly will 
bear 8 times as much as the same beam imbedded at one end, and 
loaded at the other. The calculation of the load that can be sup- 
ported by a beam (Fig. 48) firmly imbedded at both ends involves 
the discussion of the elastic curve, as the beam is bent more than 
once, and there are thus three points of rupture. I do not propose 
to give now the discussion of this curve, but will give merely the 
results. 

For a beam thus imbedded and loaded at the middle we have 

w Id 

When the load is uniformly distributed 

12KI 

w = -jt; 

The student can now refer to the table of cross sections among 
the plates with the corresponding values of I and d calculated. 
The values of K for different materials are determined experiment- 
ally, and by consulting Table IV in connection with Art. 247, 
Rankine. Thus for timber we take -fa th of the modulus of rup- 



EXAMPLES. 67 

ture for K, r or wrought iron Mb. We will take, in the following 
examples, the following values, for wood K = 1040, for cast iron 
K = 7000, and for wrought iron K = 9000. 

EXAMPLES. 

K Given a wooden beam of rectangular cross section, sup- 
ported at both ends, where b = 5 inches, h = 12 inches, / = 6 feet. 
What load will it bear with safety. 

2. The following beams of rectangular cross section are of oak, 
and all supported at both ends. 

a. b = 3 in. b. b = 5 in. c?. b = 7 in. A b = 12 in. 
/? = 8 in. A = 12 in, A = 13 in. h = 12 in. 

Z = 8 ft. I = 10 ft. 1= 6 ft. Z = 4ft. 

Find what load each one will bear with safety, both when uni- 
formly distributed over its entire length, and when applied at the 
middle of its length. 

3. The following cast iron beams, whose cross sections are hol- 
low rectangles, are firmly imbedded at both ends. 

a. I = 10 ft. b. I = 5 ft. c. I = 15 ft. 

B = 10in. ■ B = 12in. B = 10 in. 

H=10in. ' H=14id. H = 15 in. 

b = 8 in. b = 8 in. b = 7 in. 

A = 8 in. A = 10 in. h = 12 in. 

4. Determine the load each one can bear with safety, first when 
uniformly distributed throughout its whole extent, and secondly, 
when applied at the middle of its length. 

5. The floor of a warehouse which is to sustain, its own weight 
included, a load of 100 lbs. per square foot, is to be constructed of 
oaken beams 8 inches broad and 12 inches deep, find the distance 
that can be allowed between the walls, supports, etc.,; the beams 
being placed at a distance of 4 feet from middle to middle of beam. 

OT7T 

Solution. The formula is Wl = — T = |KM 2 . 



68 STRESS. 

In this case each beam has to sustain a load of 4 X 100, or 400 
lbs. per foot long measure, or - 4 T °2°- lbs. per inch. 

... W == -\%°-l .'. - 4 T ^' 2 = | X 1000 X 8 X (12) 2 

... p = 4_xi_n^_x_8 x X i L|_xj A x JJ . = 46080 gq [Bt 

.-. 1= 214.66 in. = 17.88 ft. 

5. Given in the same case b = 10 in., h = 15 in., and the load 
is one of 200 lbs. per square foot. 

6. Find the safe load of a cast iron beam, whose cross section 
is given in Fig. 49, supported at both ends. Given H = 5| in., 
h = 4.315 in., B = 1.75 in., b = 3 in., I = 6 ft. 

7. Do the same for a beam of the section shown in Fig. 50, 
Given AB = 2 in., AC = .4 in., DE == 5 in., EF = .5 in.. / = 
4 ft. 6 in. 

8. Find the safe load of a hollow cylindrical beam of cast iron 
firmly imbedded at both ends, D == 5 in., d = 4 in., I — 18 ft. 

9. Find the diameter of a cylindrical iron rod thart it may sus- 
tain a load of 200 lbs. per foot long measure. 

Explanation of the work of Eankine's, Art. 95. 
We must first observe that ffdxdy and f fdx'dy' both repre~ 
sent the area in this case of the same surface, hence we can now 
put dxdy instead of dx r dy r under the ff sign. 

x r 2 = x 2 cos 2 ft — 2xy sin ft cos ft + y 2 sin 2 ft. 

y' 2 — x 2 sin 2 ft + 2xy sin ft cos ,3 + y 2 cos 2 ft. 

x'y' = x 2 sin ft cos ft + ^y (cos 2 /5 — sin 2 /3) — y 2 sin (3 cos ft, 

Now substitute these values of x' 2 , y f 2 and x r y r in the values (2) 

of F, J' and K r , and change dx r dy r into dxdy, and we have 

r = ffx 2 cos 2 |3 ffe^/ — 2ffxy sin (3 cos /? dxdy + 

//?/ 2 sin 2 (3 dxdy, or 

F = I cos 2 ft — 2K sin [i cos /? + J sin 2 (3, and likewise 

J' = I sin 2 ft + 2K sin cos /3 -f- J cos 2 ft, and 

K 7 = (I — J) cos ft sin ft + K (cos 2 (3 — sin 2 p ! ). 

Now add the 1st two of (4) together, and we have F + J' = 

(I + J) (sin 2 ft + cos 2 ft) = I + J = //(^ 2 + y 2 )dydx. Again 

multiply them together, and we have 



STRESS. GO 

I'J' = p s i n 2 p cos 2 ^ _|_ u sin 4 /5 _ 2IK cos (3 sin 3 (3 

+ Ucos 4 (3 + 2IK cos 3 /8 sin (3 + J 2 sin 2 13 cos 2 (3— 2JK cos 3 (3 sin (3 

+ 2JK cos sin 3 (3 — 4K 2 cos 2 (3 sin 2 (3. 

= (I 2 + J 2 ) sin 2 (3 cos 2 § + IJ (cos 4 (3 + sin 4 (3) + 

2IK cos (3 sin (3 (cos f> — ■ sin (3) — 2JK cos (3 sin (3 (cos (3 — sin (3) 

— 4K 2 cos 2 jS sin 2 (3. 

Now square K' and we have 

K' 2 = (p _ 2IJ + J 2 ) sin 2 (3 cos 2 (3 + 
2 (IK — JK) cos (3 sin (3 (cos [3 — sin (3) + 
K 2 (cos 4 (3 — 2 cos 2 13 sin 2 (5 + sin 4 p). 
Now subtract, and we have 

j/j' — K' 2 = IJ (cos 4 § + sin 4 (3) + 2IJ cos 2 (3 sin 2 (3 
— 4K 2 cos 2 (3 sin 2 (3 — K 2 cos 4 (3 + 2K 2 cos 2 (3 sin 2 (3 — K 2 sin 4 (3 

== IJ(cos 4 (3 + 2 cos 2 (3 sin 2 (3 + sin 4 (3) — 
K 2 (cos 4 (3+2 cos 2 (3 sin 2 (3 + sin 4 /3) = I J — K 2 , since the paren 
theses are each equal J . 

In this equation (I' — J') 2 = (T + J') 2 — 4I 'J'> when I'J' i s 
least the subtrahend in the second member is least, and hence the 
remainder is greatest, that is (I' — J 7 ) 2 or I' — J' is greatest when 
I'J' is least, and since TJ' — K r 2 does not change value with 
the change of I',J' and K', let I'J' — K' 2 = a, now when K = 0, 
I'J' = a , when K is not then I'J' — K' 2 = a, or I'J' = a + K' 2 > 
hence I'J' is least when K' = 0, and hence, when K' = 0, V — J' 
is greatest, and when I' — J' is greatest, I' has its greatest, and J' 
its least value. 

Note on Theorem II. 

We have already seen that when K' is then I is a maximum 
and J' a minimum ; now the question arises whether it is always 
possible, to find a pair of axes for which this shall be the case. 

In order to ascertain, we make K' = in equations 4 of Rankine 
and now we find the corresponding values of I' and J', and if un- 
der any positions of the axes they become imaginary, then for those 
positions the above conditions can not be fulfilled ; but if, on the 
other hand, we find (as we shall) that they are always real, and 
never imaginary for any position of the axes, then Theorem II fol- 



PQ STRESS* 






lows. Now to test it, make IS! = in 4, and from the third 
equation we obtain 

(I _ J) C os (3 sin /? + K (cos 2 (3 — sin 2 (3) =±= 0, or 

( T — J)'. ^ . ^ ; . sin 2 (5 — 2K 
2~- sm 2p + K cos 2£ = .-.. ^-^ = j— j, or 

tan 2(3 = T j- 

Make E7 = in 6 also, and 5 and f) become 

r + ^rH J, and I'J' = IJ — K 2 . 
,-. T 2 + 2I'J' + J /2 = (I + J) 2 , and 41' J' = 4(IJ — K 2 ) ; subtract 
an d i« _ 21' J'.+ J /2 = (I + J) 2 — 4IJ + 4K 2 = (I — J) 2 + 4K 2 . 

... r — J' = VKI— J) 2 + 4K 2 |, and ,\ 
r = I±J " + yKI " J ) 2 +4K 2 ^ 



J' = 



2 ' 2 

1 + J y5(l__j)2 +4K ^ 



2 2 

Now the quantity under the radical is the sum of two squares, 
and hence is never negative ; hence follows Theorem II. 

Equations 9 are obtained directly from 4, by making K = 0. 
Note. A pair of rectangular axes which divide any plane figure 

symmetrically, are principal axes, for K' = f fxydxdy = j —rxdx ; 

i now if for every value of y there is another, equal with contrary 
sign, when the proper limits are introduced the value of X' van- 
ishes, as, for instance, in the circle, 

p V(^ 2 - x*) pr pr ( ( r 2 - x 2 ) (r 2 - x 2 ) > 

EXAMPLES. 

1. Knowing the moment of inertia of an ellipse about the 

7rab s . -a s b 

major axis to be — j-,and about the minor -j~ , find its moment of 

inertia about a pair of axes, making an angle § — 30° with these. 
Solution. 

r = I cos 2 [3 + J sin 2 (3 = ^J- cos 2 30° + ~ sin 2 30°. * 



CONJUGATE AXES. 



71 



~a*b -ah 3 

J' = I sin 2 B + J cos 2 S = -p- sin' 2 30° + 1 — cos 2 30°. 

K' = (I — J) cos 30° sin 30° = --— - ^"— - cos 30° sin 30% 

or I' = ^(a 2 ^/3 + J 2 ), J' = ~- (a 2 + &V8), 

K' = ^- (a 2 - 6*)V3. 

If the ellipse becomes a circle, a = b and T = J', and K' = 0. 



CONJUGATE AXES. 

We have already seen that the axis conjugate to any neutral 
axis is the line joining the centre of gravity with the centre of 
stress for that neutral axis. So that as soon as a neutral axis is 
given its conjugate is at once determined. 

We have by Art. 94 (Rankine) the coordinates of the centre of 
stress, and hence we have the direction of the conjugate axis, for 

cot 6 = — = y ; where d is the angle between the neutral and 

conjugate axis if K = 0, cot = 0. and d = 90° .\ when we take 
one of the principal axes as neutral axis the other becomes its 
conjugate axis. 

Theorem IV. We wish to prove that the neutral and conjugate 
axes are interchangeable. Now to do this we have only to take the 
conjugate axis as neutral axis, and find the angle between it and its. 

conjugate axis ; to do this we have cot d = -jr, in which we must 

substitute the values of K' and I' from 4, substituting for £?. 

K I 

Now we have cot 6 = j .\ tan z= jF' 

tan I 



and cos = 



J (1 + tan 2 0) ~~ V(I 2 + K 2 )' 

1 K 

V(l + tan 2 0) ~ V(I 2 + K 2 )" 



INTERNAL STKKSS. 



Now substitute in the 1st and last of 4 and we have, 
IK 2 -f Jt 2 — 2lK 2 I(IJ — K 2 ) 



K' = 



y — 

I 2 + K 2 

p K — UK + K 3 — I 2 K 



I 2 +K 2 ' 
K (IJ — K 2 ) 



I 2 + K 2 
K' 



I 2 + K 2 
K' K 

... cot 0' = ~y = — y /. 0' = — /. Q. E. D. 

Proof of formulae 13, page 81, Rankine. 






*■; 






yy 



Equation of tangent is — j- + -p- = 1, .-. OT = n - 

1 ab ab ab 

' OY' = 7 



v/'(S + ft " (£ + ¥) 



2 H 



but from the equation of ellipse (pole at centre), 

6 = — : j .:n= jr = (a 2 cos 2 p + 6> 2 sm 2 p) 2 * 

(a 2 cos 2 (3 + 6 2 sin 2 (3) 2 ° 

II. ^ = 0C 2 _ Q T i = a '2 _ n 2 ? but a f2 _|_ V 2 = a 2 + ^ 

... a ' 2 = a 2 + b 2 — U 2 .-. * 2 = a 2 + 6 2 — V 2 — ri 2 = 

a 2 b 2 

a 2 + b 2 — v2 ,^ 2 K > _l A2 ^,,2 / .> — a2 cos2 f> — b 2 sm ' 2 P* 



a 2 6 2 



a 2 cos 2 (3 + 6 2 sin 2 (3 

= a 2 sin 2 8 + 5 2 cos 2 (3 — -t, 2 , r 7 -> . 77 T 7 

r ' r <sr cos 2 p + tr snrp 

(a 4 + £ 4 ) sin 2 ft cos 2 (3 + a 2 5 2 (sin 4 (3 + cos 4 (3 — 1) . 



but sin 4 (3 + 2 sin 2 (3 cos 2 (3 + cos 4 (3=1, 
sin 4 (3 + cos 4 (3 — 1 = — 2 sin 2 (3 cos 2 (3 



A * 2 = 



(a 4 — 2a 2 b 2 + V) sin 2 (3 cos 2 (3 



,-. r^ = (a 2 — b 2 ) sin (3 cos £># 



INTERNAL STRESS. 

Examples on Arts. 96 — 99. 

1. A rod of elliptic cross section (semiaxes 1.2 and .7 inches 
respectively) has suspended to it a weight of 2,000 lbs. What is 



COMPOUND STRESSES. 73 

the intensity of the normal, and what the tangential stress on a 
plane inclined at an angle of 60° to the axis of the rod. 

Solution 1st. Area of normal cross section = ~ab = tt(1.2)(.7) 
= 2.638944 sq. inches. 

Area of oblique section = 2.638944 -f- cos 30° = 3.04717. 

Total normal component = 2,000 cos 30° = 1732.06 lbs. 

Total tangential component = 2,000 sin 30° = 1000 lbs. 

Normal intensity = ^, 7 o4 2 fi7 = 568.41. 

Tangential intensity = rffi&fcj = 328.17. 
Solution 2d. By formulas 

Pt Z=Z^ x COS 2 6 = 2.6 2 3°8°9°4 4 cos 2 30° == 568.41. 

Pt =P* cos sin = gillie cos 30° sin 30° = 328.17. 

2. Find the intensity of the stress on a normal cross section of 
each of the following struts and ties. Each frame is loaded with a 
weight of 1,500 lbs. 

The cross sections of all the above beams are rectangles, whose 
sides are 2 and 5 respectively. 

It is necessary to notice that the whole stress, which is the re- 
sultant of all the stresses, is a single force acting at a single point ; 
but the intensity of the stress is the amount per unit of area. 

COMPOUND STRESSES. 

We have already seen that stress consists of a distributed force 
acting on a body, and counteracted by the cohesion, etc., of the 
body ; thus a body in a state of strain is in equilibrium, and all our 
problems of stress, are merely problems of equilibrium, and are 
solved by imposing the conditions of equilibrium. Moreover, a 
body may be acted on by stresses in different directions, and in- 
deed practically always is, though in many cases we need only 
consider the stress in one direction ; but, for instance, suppose we 
have a lattice supporting a weight, and that the two cross bars are 
pinned together, as in the figure, there is a stress acting in the 
direction of each lattice on the pins. So suppose a vessel contain- 
ing water, or any liquid, a stress is exerted in all directions, and 

10 



/4 COMPOUND STRESSES. 

we must be able, knowing the load, etc., to determine the stress on 
any given surface. 

Proof of Kankine's Art., 101. 

Here we take a prismatic portion of the body, the sides of whose 
base are parallel to the direction of the 1st stress, and also the 
plane on which it acts. Now the forces on the plane AB are 
counteracted of course by a set of forces, and the resultants of each 
of these sets act through O, and in the same straight line ; hence 
the forces acting on the planes AD and BC must be balanced en- 
tirely, independent of any of the forces on AB or DC. Now sup- 
pose the stress on AD and BC did not act in a direction parallel to 
AB, we should have their resultants equal, but not .directly op- 
posed ; hence we should have a couple, and there is no opposite 
couple furnished by the other stress ; hence the body would not 
then be in equilibrium. .-. Q. E. D. 

This condition of the stress is exhibited in the preceding exam- 
ple of the lattice. We may have three conjugate stresses acting on 
a body, as, for instance, a gusset in a corner of an iron T-shaped- 
girder. 

Note showing the work of Art. 102. • 

Let x, y, z be the directions of the three stresses. I have not 
drawn the perpendiculars, not to complicate the figure; we 
have from spherical trigonometry cos yz = cos zx cos ay + 
sin zx sin xy cos (ZXY). Now the angle ZXY is the angle be- 
tween the planes ZX and XY, and is the supplement of the angle 
between two lines drawn perpendicular to these planes ; hence 
ZXY = 180° — vw .-. cos ZXY = — cos vw .\ 

cos yz = cos zx cos xy — sin zx sin xy cos vw 
cos zx cos xy — cos yz 

/. COS VW = ; -. — • 

sin zx sin xy 
From the last we wish to find sin vw. 

. . OH / C0S ZX C0S X V C0S V Z \ Z 

snr vw = 1 — cos" 1 vw = 1 — I ; —. — ■ I 

\ sin zx sin xy / 

cos 2 zx cos 2 xy — 2 cos zx cos xy cos yz + cos 2 yz 

sin 2 zx sin 2 xy 



COMPOUND STRESSES. 75 

sin 2 zx sin 2 xy - cos 2 zx cos 2 xy -\- 2cos zx cos xy cos yz - cos 2 yz 



(1- 


sin 2 ## sin 2 xy 

- cos 2 £a') ( 1 - cos 2 cn/)- cos^a* cos 2 cn/ -f- 2cosz# cosxy cosyz - cos 2 yz 




sin 2 Ze£ sin 2 #y 
1 - cos 2 £03 - cos 2 o:y — cos 2 yz + 2 cos fry cos zx cos yz 



sin 2 #y sin 2 zx 

.'. sin wo ==— ; , and so far wu and itv. 

sin xy sin zar 

Pass the plane XOA perpendicular to ZY, and we have a right- 
angled triangle, XAY, where XA is the measure of the angle he- 
sin XA 
tween X and the plane ZY. Now we have sin X YA = — — - — ; 
r sin xy 

but XA — tix — 90°, and XYA = inc. 

cos ux 

.-. sm inc = .-. cos ux = sin #y sin ww. 

sm cry * 

Now substitute the value of sin inc just found, and we have 

Vc 

cos ux — —. , and so for the other. 

sin yz' 

Restricted case. I. In this case xy = 90° .*. cos xy = 0, sin xy 
= 1,/C = 1 — cos 2 yz — cos 2 zx. Make these substitutions in 
the general formulae. 

Restricted case. IL Here yz = 90°, zx = 90° ,\ cos yz = 
cos zx == 0, sin yz — sin zx = 1, C = 1 — cos 2 xy = sin 2 #y. 

Now substitute. 

EXAMPLES. 

1. Given three conjugate stresses, such that&y= 30°, zx = 40°? 
zy = 60°, find uv, uw, vw, ux, vy, and icz, respectively. 
Solution. 

C = 1 — cos 2 yz — cos 2 zx — cos 2 xy -f- 2 cos yz cos zee cos #y 
= 1 _ cos 2 60° — cos 2 30° — cos 2 40° + 2 cos 60° cos 30° cos 40°. 
cos 60° = i = sin 30°, sin 60° = *V3 = cos 30°, cos 40° = 
.76604, sin 40° = .64279. 

C = l — i — | — .586815816 + i*/3 cos 40° = 
.076592815816; J C =.27676. 



COMrOUND STRESSES. 

cos 40° cos 30° — cos 60° .76604 X .86603 



cos vw 







.64279 X 


.5 


.5 


X 


.86603 — 


.76604 






.5 X .86603 


.5 


X 


.76604 — 


■ .86603 



sin 40° sin 30° 

= .50845. 

_ cos 30° cos 60° — cos 40° _ 
cos uw — sin 30 o sin 6Q o 

= _ .76906. 
cos 40° cos 60° — cos 30° _ 
cos wo — sin 40° sin 60° — "" .64279 X .86603 

= —.86767. 
sin W0 = . 86112 vw= 59° 26' 22" 

sin uw = .63915 uw = 140° 16' 10" 

sin uv = .49712 uv = 150° 11' 30" 

cos ux = .31957 ux = 71° 21' 48" 

cos vy = .43056 vy = 64° 29' 49" 

cos wz = .55352 wz = 56° 23' 27" 

2. Given ^y .= 25°, yz = 45°, zx = 45°, find as above. 
Case I. xy = 90°, sb = 50°, ^ = 60°. 

Case II. ^y = 30 °> ^ = 90 °^ X V = 90 °- 

Now Articles 103 and 104 of Rankine can be read. 

Note on Art. 103 of Rankine. 

We must bear in mind that stress is a set of forces balanced by 
another equal and opposite set of forces. Now a shearing stress 
is equivalent to a couple, for it tends to draw a part of the fibres 
one way, and the fibres adjacent to it in the opposite direction, 
hence being a couple it can only be balanced by a couple of equal 
moment and opposite sense. Keeping this in mind, Eankine's 
demonstration is clear. 

Note, on Art. 104 of Rankine. 

The demonstration is clear, only it is necessary to pay attention 
to the notation. When we have a body referred to three rectangu- 
lar axes Oa?, Oy and Oz, we call the x plane the plane perpendicu- 
lar to the axis Ox, or, which is the same thing, the plane ZOY, etc. ? 
then p y21 for instance, means the intensity of the stress on the y 
plane, in the direction Oz. In Art. 105 of Rankine, we must find 
the total stresses on OAB, OAC and OBC, respectively, and lay 



COMPOUND STRESSES. i i 

off OF, OE and OD, to represent these, then by oompoumTiug 
these we obtain the total stress on ABC, then dividing this by tie 
area of ABC we have the intensity of the stress on ABC. 

The object of Art. 106 is. knowing the direction and intensity of 
the stress on three given rectangular coordinate planes, to deter- 
mine the stress on any other given set of rectangular planes. 

To do this, the resultant stress is first determined on any one 
plane. ABC. Now resolve this stress into three components parallel 
to the axes of x. y and z, and here we must notice the notation, viz. : 
p nx denotes the intensity of the stress on ABC in the direction 
Ox; y> DV that on ABC in the direction Oy: and p nz that on ABC in 
in the direction Oz ; p vx , that on ABC in the direction Ox', etc, 
Now the component of the stress in the direction Ox' will be found 
by projecting the resultant stress on Ox', this is p^^. and we obtain 
the same result by projecting the three components p nx . p n . and 
p nz . on Ox' ; we thus obtain jo m , = p nx cos xx r -f- p ny cos yx + 
p nz cos zx'. and so the values given in the text of p nj , andp nz ,. 

VTe have thus found expressions for the stress on any oblique 

plane in any given direction, and we may now suppose the oblique 

plane to be that perpendicular to Ox', and we shall have j» x , x , = 

Pnx cos :r *' + Puj cos y :ff + i?nz ( ' os * x % and imposing the same 

condition on the equations 1 of Art. 105. we obtain 

P** = V*x cos xx' + p xy cos yx' + p zx cos zx'. 
p^ T = p XJ cos xx' + Py 7 cos yx + p JZ cos zx'. 
Pnz =?zx cos xx +p JZ cos yx' + p zx cos zx'. 
Substitute these values of p^. p nj andy nz . and we have 

P*>* = (Rxx cos xx' + p^x cos yx -\- p zx cos zx) cos xx' 

+ (p xy cos xx' + Pjj cos yx' + p yz cos zx) cos yx' 

+ {p*x cos xx' -f- p yz cos yx' + p zx cos zx') cos zx' 

= p xx cos 2 xx 9 + Pjj cos 2 yx' + p zz cos 2 zx -\- 2p yz cos yx cos sa/ 

~T '2px 7 cos skb' cos yx' + 2jt? zc cos za/ cos xx'. 

By making the same substitution in the other two equations we 

obtain the values of p x , y , and p x , z , ; then suppose the oblique plane 

to be. first, that normal to Oy'. and second, that normal to Oz', and 

we obtain the remaining values. 



78 COMPOUND STRESSES. 






We have thus determined the stress on that of the three new 
coordinate planes, x'Oy', x'Oz and y f Oz\ and now we wish to as- 
certain whether there can always be found for any state of stress 
in a body three planes at right angles to each other, on which the 
stress is wholly normal, or in other words, three principal axes of 
stress. 

To do this we refer back to Art. 105. and impose the condition 
that the stress shall be wholly normal, or that xr shall equal xn, 
yr = yn, and zr = zn .*. p nx — p r cos xn, p ny — p r cos yn, and 
p nz = p r cos zn. Substituting these in equations 1, we have equa- 
tions 2 of this article, from which we must eliminate the cosines. 
Thus by the method of Undetermined Multipliers, multiply the # 
2d equation by /, and the third by m, and add all three together, 
and we have 

cos xn (p xx — p r + lp xy + mp zx ) + 
cos yn \p xy + I (p yy — p r ) + mp JZ \ + 
cos zn\p zx + lp yz +.m(p zz — p r )\ = ; 
and as we can impose two conditions make 

^ xx — Pr + ¥*y + m V™ = 0, and p xy + l(p yy — p r ) + mp yz = 0, 
and hence follows p zx + lp yz + m(p zz — p r )- 

Now from the first two find I and m, and we have 

7 __ (Pt—P**)Py* +P*yP™ OTwl _ _ ( Pr-Pxx)(Pr-Pyy) ~ P*y* 

l — l / \ « a n ci n o — i / \ 

PxyPjz + PzAPr — Pyy) PxyPyz + PzAPr ~ Pyy) ' 

Substitute these in the third, and clearing fractions, we obtain 

P*x\P*jPjz +Pz*(Pr —Pyy)\ +Pyz\(Pr — Pxx)P 7 z + PxyPz*\ 
+ (Pzz —Pr)\(Pr —Ptx) (jPr ~ Pyy) ~ P^\ = 0. 

Whence by performing the operations indicated, we obtain the 
cubic p r s — Ap T 2 + B/? r — C == 0. This gives three values of p T , 
and hence there are for every condition of stress in a body three 
planes at right angles to each other, on which the stress is wholly 
normal. 

Equations 5 may be obtained by equating the value of cos zn in 
two of equations 2, and thus obtaining a relation between cos xn 
and cos yn. 



STRESS PARALLEL TO ONE PLANE. . 79 

To obtain equations 7 we have the whole stress on ABC equal 
to p • Area ABC. and that on OBC equal to p } • Area OBC ; pro- 
ject the 1st on Ox and we have 

p - Area ABC * cos xp = p l • Area OBC, or 
p • Area ABC cos xp = p 1 • Area ABC cos xn /. 
p> cos xp = Pj cos xn, and so 
p cos yp —p 2 cos yn, and p cos zp — jt? 3 cos zn. 
To obtain equation 11 project first the whole stress on ABC on 
the normal On and we obtain p • Area ABC cos np, and then 
project each of the components of the stress in the same manner, 
and add their projections and we have 

p • Area ABC cos np = (p • Area ABC cos xp) cos xn + 
1 (p - Area ABC cos yp) cos yn + (p ■ Area ABC cos zp) cos zn. 
.-. cos np = cos xn cos xp -\- cos yn cos y_p -f~ cos zn cos 2p, and 
the rest of the reduction is easy. 

STRESS PARALLEL TO ONE PLANE. 

In practical problems we have to do generally with stress par- 
allel to one plane. Such is the case in lattices, in earthwork, and 
all kinds of structures. 

The object in the following problems is, Given the conditions of 
stress in a body, that is, Given the stress on any two planes, both 
as to direction and intensity, to deteimine that on any given plane 
whatever. So that we may be able to know what resistance the 
body must oppose in any given direction, and hence what amount 
of strength is required. 

Always now it is implied that the stress is parallel to one plane 
(the plane of the paper in the figures). 

Note on Problems I and II, Rankine, page 96. 

We must notice that in finding the resultant stress on the plane 
AB it would not do to compound merely the intensities of the 
stress, as we do forces. The force acting on OA is not p y but 
p y - OA, and that on OB is p x • OB, so when we have OR for 
the resultant force this is the. force on AB, and the intensity is 
OR 
AB = P' 



80 EXAMPLES. 



EXAMPLES. 






1. Given two conjugate stresses of intensities, 4 and 5, respect- 
ively (4 and 5 lbs. per square inch), p x = 4, p y = 5, their direc- 
tions inclined to each other at an angle 60°, what is the intensity 
of the stress on a plane at right angles to the axis of X. 

Solution. In this case OAB becomes a right triangle, right- 
angled at A, and ABO = 30°. Now take OA = 1 (inch say), 

OA 

then OB = wtto = 2, and AB = OA tan 60 = V3 .-. 

cos 60 

OD = 4, OB = 8, OE = 5, OA = 5, and 

OR = V(8 2 + 5 2 + 2-8-5 cos 60°) = V(64 + 25 + 40) = Vl29 

11.35 11.35V3 
= 11.35 /. p T = - y. r — o = 6.55273 (pounds per 

square inch say). 

2. Given p x = 7, p y = 9, AOB = 45°, OAB = 30°. 

3. Given p x = 5, p y = —8, AOB === 30°, OAB = 100°. 

4. Given p x = 8, p y = 10, AOB = 75°, OAB = 90°. 

5. Given p x = —3, p y = 5, AOB = 90°, OAB = 45°. 

6. Given the stress, P^ig. 51, on the plane OA of intensity p y 
= 7, and on OB intensity p x = 5, and angle the 1st makes = 30° 
with OX, angle xn = 30°. 

Solution. 

p xx = 5 cos 30° = 4.33015. 

p yy = V^7 2 — (2.5) 2 ^ = V(49 — 6.25) = V(42.75) = 6.5383. 
. •'•Pr = 

«/ \Pxx 2 COs2 Xn +i 9 yy 2 Sm2 Xn +iPxy 2 + 2Pxy(Pxx~\-Pyy) C0S Xn Sm ^ 

^/ } (4.33015) 2 cos 2 30°+(42. 75)sin 2 30°-f 6.25+5(1 0.86845)cos30°sin30° } 
= V(p n 2 + ^ r 2 ) = from what follows, V(728.880697) = 26.997. 
p u = (5 cos 30°) cos 2 30° + (42.75) sin 2 30° + 5 cos 30° sin 30° 
= W + 10 - 68 + 2.16507 = 26.9075. 



EXAMPLES. 81 

p t = (2.20815) cos 30° sin 30° + (2.5) (| — i) =.95616 + 1.25 

= 2.20616. 
p t 2.20616 



tan nr = 



p n "26.9075 logp t = 0.3436370 

log^ n = 1.4298733 



log tan nr = 8.9137637 
nr = 4° 41' 14". 

7. Given p 7 = 8, p x = 6. Angle p y makes with OX = 45°, 
and xn = 40°. 

8. p v = 7,p x = 5. Angle p y makes with OX = 75°, xn = 45°. 
Art. 109 of Rankine we have 

cos xn sin xn p xy . 

— - -— . . j^ sm 2xn = 2 cos cm sin xn. 

cos' 2 cm — sin" 5 cm p xx — j? yy 

sin 2xn 2p xy 



and cos 2x?i 



cos 2^'^ j(? xx — p y 
or tan 2x/i 



2/> 



hence follows what Rankine says about the principal axes. 

Arts. 110 and 111 are clearly given in Rankine, but I wish to 
comment a little on their results. From Art. 110 we see that two 
principal stresses of the same kind and of equal intensity produce 
on any plane (parallel of course to the stresses) a stress of the 
same intensity, and normal to the plane of action. This is the 
case in a perfect fluid, and it is borne out by experiment. If, for 
instance, we have a vessel of any shape whatever containing a 
liquid then at any given depth below the free upper level, the pres- 
sure is the same in all directions, and always normal to the plane of 
action as we can see by tapping the vessel at any point. 

If, on the other hand, we have the two stresses of opposite kinds, 
(one a pull and the other a thrust), then the resultant stress on any 
plane is still of the same intensity, but it is no longer normal to 
the plane of action, but makes with the axes of principal stress, 
angles equal to those made by the normal to the plane of action. 
(Fig. 52.) 

Thus if OX and OY represent the axes of principal stress, the- 
11 



82 ELLIPSE OF STRESS. 

stress on OX being a pull, and that on OY a thrust, and if we wish 
to find the stress on the plane AB (or any plane parallel to it,) we 
should find by the construction of Art. Ill, OR, where XOR = 
XOC, and if we wish the stress on DC by the same construction we 
obtain OP. If, now, COY = BOY = 45°, then the lines OR and 
OP would coincide with AB and CD respectively, hence follows his 
conclusion. As an example of this condition of a pull and a thrust, 
we can take an iron bracket, on which the stress (when loaded) is 
evidently a pull in a horizontal direction, and a thrust in the verti- 
cal. 

ELLIPSE OF STRESS. 

Before entering on the Problem of the Ellipse of Stress, we 
must remark that its object is a simplification and also a generaliza- 
tion of Problem II, page 97 (Rankine). The practical engineer 
often finds it much more convenient to lay off certain lines to rep- 
resent the different quantities in his calculations, and then to make 
a geometrical construction, which enables him by simply measuring 
a certain line to ascertain the result of his calculation, especially 
if he can make one figure serve him instead of a number of calcu- 
lations, rather than to make a lengthy numerical calculation every 
time, in which also he has a great many chances for error. 

Suppose we have two principal stresses, p x and p J9 now 

p x — — g 1 2 > and Pj — 2 ~~ 2 J 

hence we may imagine, instead of the two original stresses, p x and 

p y , the 4 stresses, viz. : 

P* H~ Py 
1st. A stress of intensity g acting along OX. 

2d. " " " " 

3d. " " " 

4th. u ■ " " " 





2 




^x 




Py 




2 




Px 


+ 


Py 




2 




Px 




Py 



ELLIPSE OF STRESS. 83 

1. Now let us find the stress on the plane AB resulting from 
the 1st and 3d. Since these are of the same kind and of equal 

V ~f~ V 
intensity, viz, ^ — -, they produce a stress of the same intensity, 

acting in the direction ON perpendicular to AB. Lay off (on any 

P ~\~ V 
scale) OM == 9 — * (say an inch to a pound), and this repre- 
sents the resultant intensity of the stress on AB from the action 
of the 1st and 3d. 

Now take the 2d and 4th. These are two stresses of equal in- 
tensity and opposite kind, hence by Art 112. they produce on AB 
a stress of the same intensity, but whose direction makes with OX 
an angle equal to that which ON makes with it. If we take P, • 
such that MP = OM, POM is an isosceles triangle, and MPO = 
MO P, hence PM will be the direction of the resultant stress. Now 

on PM lay off MR = ~ — ? (on the same scale), and join OR. 

(We already know that OR will represent the resultant of two 
forces acting at the same point, represented in direction and mag- 
nitude by OM and MR respectively, and since it is altogether stress 
on AB we are considering, we may combine the intensities just as 
we do the whole stress.) Hence OR represents the resultant in- 
tensity and direction of the stress on AB. 

Two things remain to be done. One is to express algebraically 
the magnitude and direction of OB, and second to see if we cannot 
find some line straight or curved in which the point R must always 
fall for every position of the plane AB. Now to find the algebraical 
value of OR, we have 

PM == OM = QM, but OM = Px 2 J , hence 
PA = 2 • P -^ll =p x + Pj , and RM = P^Zli 

.-. QR = QM + MR = p x , PR == PM — MR = p r 
Now project RM on OX and PR on OY, and the projections 
will be p x cos xn, and p y sin xn, respectively, whence OR 2 = 



84 ELLIPSE OF STRESS. 

(p x cos xn) 2 -f~ (Pj & m xn) 2 , since we shall have a right angled 
triangle, with OR for hypothenuse, and these projections for sides. 
.*. p r = */{p 2 cos 2 xn + Pj 2 sin 2 xn). 
2. From the triangle EOM we have 

sin ROM : sin RMO : : RM : OR : : P *~ Pj : p r . 
.-. sin ROM = sin RMO ■ Px I~ Py , but EOM = nr, and ROM 

"Pv 

= MQO + MOQ = 2MOQ = 2(90 — xn) = 180° — 2xn. 

/. sm wr = sin Ixn— ~ • 

2p r 

We have thus found the algebraic expressions for the intensity 
of the resultant stress and its inclination, and now we proceed to 
find the locus of the point E.. 

First the locus of M is a circle, for in whatever position the 

P ~\~ P 
plane be, OM = -^-~ — -, and this does not change when AB 

changes, hence M is always at the same distance from O, and .*. 

the locus of M is a circle. 

Now for the locus of E we have seen that RA = p x and 

PR = p y .'. PA =jp x ~\~ Pr Draw perpendiculars from E to the 

axes of X and Y, so as to obtain the coordinates of E, and we 

shall have X = QE cos xn = p x cos xn, y = PE sin xn == 

p 7 sin xn, these evidently change for every value of xn, but we 

have 

x y 

— = cos xn, — = sin xn. Now square and add, and we have 

Px Pj 

x^ v 2 x y 2 

— 5 + — a = cos 2 xn + sin 2 xn, or — o + — 5 — 1? where xn has 

Pj Py 2 Pj Py 2 

disappeared, and we have the equation of an ellipse whose semi- 
axes are p x and p y , respectively ; hence E is always situated on 
this ellipse. 

Hence instead of the preceding construction for OE we can 
construct our ellipse, then draw the perpendicular ON to the plane 

AB, and then lay off OM = 1 ^^ 1 , and then draw PM = OM, 
and where it cuts the ellipse will be E. 



EXAMPLES. 85 

EXAMPLE. 

It can easily be shown that in a boiler or tank for holding water, 
the intensity of the pressure in a vertical direction is double that, 
in a horizontal direction. Now suppose we want the intensity of 
the pressure on any oblique surface. Construct an ellipse with 
semiaxes 2 and 1 (inches), respectively, draw AB (Fig. 53) repre- 
senting the plane on which we wish to know the intensity of the 
stress, draw ON perpendicular to it, lay off on ON a distance -^-J- 1 
= J, and draw PM = OM, then will OR represent the intensity 
of the stress on the oblique plane. 

The intensity and direction being found we can decompose it 
into a normal and tangential component. The normal component 
will be the projection of OR on ON, or which is the same, the 
projection of the coordinates of OR, viz., p x cos xn and p 7 sin xn 
on ON .*. p n = p x cos 2 xn + p Y s^ 2 xn ' 

The tangential component will be found by projecting the same 
on AB .*. p t = p x cos xn sin xn — p y sin xn cos xn. 

Problem II The shear is the projection of OR or MR on AB 
.-. is less than MR, except when MR is parallel to AB, and repre- 
sents the shear. 

EXAMPLES. 

1. Given two principal stresses of intensities, p x == 16. p 7 — 9, 
respectively. Find the intensity and direction of the stress on a 
plane where xn = 35°. Also find the normal and tangential com- 
ponents. 

Solution. OM = Px + 2 Pj = %!, MR = P * ~ Pj = J, 

or = Pr — VK 16 ) 2 co§2 35 ° + ( 9 ) 2 sin2 45 °l = 

V(198.4262362884) == 14.08638. 



sin nr = — g- — sin 2xn = o^.tt 2 t 6 



86 EXAMPLES. 

p n = p x cos 2 xn + p Y sin 2 xn log 7 = 0.8450980 

=10.73610 + 2.96094 = 13.69704 log sin 70° = 9.9729858 

0.8180838 
log 28.17276 = 1.4498294 
Pt == (Px — Py) cos xn sm xn ^°S sm nr 9.3682544 

— P* ~ p ? sin 2xn = 3.28891. nr = 13° 30' 8". 

2. Given p x = 23.134. p y = 17.421. 

1st. xn = 41 . 2d. aw = 87° 3d. xn = 53°. 

3. Given p x = 13.427. Pj = 14.329. am = 75°. 

4. Given ^ = 12.727. p y = 5.432. xn = 45°. 

5. Given j» x = 32.531. jt? y — 17.853. aw — 33°. 

Find in each of the above, jp r , ^ n , p t , nr, and also construct the 
figure in accordance with the data. 
Also the following : 






6. 


p x = 10.451. 


Pj = - 


—8.634. 


xn == 23°. 


7. 


p x = 9.427. 


Py = ' 


—3.231. 


xn = 135°. 


8. 


p x = 8.372. 


Py = - 


-7.438. 


xn — 127°. 


9. 


p x = 7.241. 


Pt~ 


5.327. 


a:?i = 475°. 


10. 


^ x = 12.458. 


Py^ 


6.453. 


xn — 37° 5' 22 



Problem IIL 

Case 1. BM < OM .-. MOR < 90°, hence MOR is a max 
imum when sin MOR is a maximum ; but 

sin MOR : sin ORM : : MR : OM, or 

sin MOR = p *~ p * s in ORM. 

Px+Py 

Hence sin MOR is greatest when sin ORM is greatest ; but the 
greatest value of sin ORM is when ORM is 90° ; hence MOR = 
nr is a max when MR is perpendicular to OR, and then sin maximum 

Px — Py . _, Px — Py 

xn = nr = \ , or max nr = sin i , — — • 

p x + p 7 ' p* + p Y 

2. PMN = OMQ = MOP + MPO = 2MOP == 2xn .-. 

90° -f~ max nr 
xn = ^PMN, hence follows xn — ~ 



EXAMPLES. 87 

3. Pv = V(OM 2 — MR 2 ) = V(OM + MR)(OM — MR) = 
== o/p*p 7 . 

In Case 2d p T = V(MR 2 - OM 2 ) == V(MR - OM) (MR + OM) 

V-p xPy . 

EXAMPLE. 

1. In the 1st of the above examples find the plane for which 

the tangential stress is greatest. Also find the position of the 

plane on which the stress is most oblique, and in each case the 

intensity of the stress. 

p . — p 
Solution. 1st. maxj? t = ~ — - = f, and the plane makes 

45° with Ox. log 7 = 0.8450980 

logr-25 = 1.3979400 



16° 15 2 36". log sin nr = 9.4471580 

90° + 16° 15' 36" Q , „ 

Corresponding value of xn = ~ =58 7 48 . 

Intensity = VjP x jf? y = 4*3 = 12. 

2. Do the same with each of the above examples. 

Problem IV. 

This is the converse of the ellipse of stress ; there we had given 
the two principal stresses p x and p y ,[ and we were required to find 
the resultant stress on any given plane AB. 

Suppose, now, we should by Problem I find the stress on AB 
(construct it), then find it on another plane A'B', then suppose the 
principal axes erased, and that we have given us merely these two 
planes and the respective stresses on them, and are required to find 
the axes of principal stress, and also its intensity. This is Prob- 
lem IV. 

Now we know that in Problem I, to find R we first lay off on 

V ~\~ V - 
OX a length representing ^ u 3 then on a line that makes equal 

p . — p 
angles with x to what ON does, a distance ~ ~j and we thus 

have R. Do the same for the other and we have two triangles, 



8S ELLIPSE OF STRESS. 

where OM = OM' = P ^^ Pj , and MR = M'R' = Px ~ Py ,and 

the angles are respectively NOR and N'OR', the question then 
resulves itself into this, viz. — 

Given OR, NOR, OR', NOR 7 to construct two triangles, with 
this side and angle, respectively, whose other two sides shall be 
respectively equal to each other (one in each triangle). This is 
done by the construction in the text for MR' = MR, and OM is 
common, hence OMR and OM'R' = OMR', are the two required 

triangles, and OM represents Px 2 Pj , and MR = MR 7 = P *'Z Py ' 

Half their sum will be p x , and half their difference p T 

Moreover, from Problem I we know that xn = JNMR (the tri- 
angle ORM in either figure is the same), and xn r = JNMR', we 
thus can construct the axes of principal stress and obtain their 
intensities by measuring OM and MR. 

Fig. 54 shows the construction when the stresses are of different 
kinds, and in this case, by referring to the second figure of Prob- 
lem I, 2xn = RMO .-. xn = iRMO, xn r = *R'MO. 

The figures now give at once in the 

1st case, xn + xn' = -i-NMR + iNMR' = NMS. 

2d case, xn — xn' = iRMO — iR'MO = RMS. 

So far we have made geometrical constructions, and now we 
must express the quantities, viz., the intensities of the principal 
stresses, and their directions. 

First, to find OM. From the triangle OMR, we have 
MR 2 = OM 2 + OR 2 — 20MOR cos ROM ; 
and from OMR', MR' 2 = OM 2 + OR' 2 — 20MOR cos R'OM ; 
but OR = p, and ROM = nr: OR' =p', and R'OM = nV, 
hence the equations become 

MR 2 = OM 2 +p 2 — 2p-OM cos nr, and 

MR 2 = OM 2 + p' 2 — 2/0 M cos nV ; subtract and 

= p 2 — p' 2 — 20M {p cos nr — p' cos wV), or 

nAf _ f —p' 2 : _ Px+Pt 

2 (v cos nr — p cos nr) I 



EXAMPLES. 89 

Now substitute this for OM in the 1st and 2d equations, and we 
have 

/ f /» -f- p \ 2 ) 

y J y 2 / + ^ 2 — (Px + ^ c os fir j- 

= v { (^4^) 2 +^ /2 - (p- + p^p' cos wV } 

Now draw a perpendicular from R to OM, and we have 
OR cos NOR = OM + MR cos NMR (by projection), or 

^? cos wr = — ~ — - + o cos ^ ra » 

2p cos rcr — ^ x — p x , , . ^^, 

/. cos 2x?z = : — — , and by proiectmff OK we 

, r r, P*+ Pv i P* — Px „ / 

have, p cos nr = — ~— ^ — f- — ~ — - cos 2xn . 
2p' cos nV — p x —Pj 

/. COS 2aw = — — • 

Px— Py 



EXAMPLES. 



1. Given OR = 17. OR' = 12. nr = 37°. «V = 23° 

/>x + Pv _ mr _ 289 — 144 145 

2 ~~ um ~ 2(17 



(17 cos 37° — 12 cos 23°) ~~ 5.06176 
= 28.646. 

Rx ~ Fv = V $820.593316 + 289 — 2(28.646) (17) cos 37°^ 

= V(720.67001152) = 26.845. 
.-. p x = 28.646 + 26.845 = 55,491. 
Vj = 28.646 — 26.845 = 1.801. 
2p cos nr — (p x + p v ) 27.15376 — 2(28.646) 

cos 2xn — - — p ^_ p ^ ~ — 2(26.845) 

= —56133 /. 2xn = 124° 8' 52" .-. xn = 62° 4' 26°. 

22.092 — 57.292 
cos zxn = 9f9fi ai^ == — -65747. 

2xri = 131° 6' 24" .-. xn' = £5° 33' 12". 

12 



DO EXAMPLES. 

Case 3. In this case OR and OR' coincide in direction, and 

hence we have (Fig. 55) 

nri = iNMR + JNMR' = NMS == MOS + OSM = nr + 90°. 

OR -I- OR' v 4- »' 
Also OS = OM cos MOS = OM cos nr = -^ = L -j JL 

. hat— P*+Pj P+p\ 

.'. UAL — ^ — rr ' 

A I cos wr 

and OM 2 = OS 2 + MS 2 . MR 2 = SR 2 + MS 2 ; subtract, and 

OM 2 — MR 2 = OS 2 — SR 2 = (OS + SR) (OS — SR) = 

OR • OR' = pp' .-. MR 2 = OM 2 

- pp> or —^- = vH-r) -pp \ = 

V 1 4 cos 2 nr PP ) 

Case 4. In this case the figure becomes Fig. 56, and we have 

by projections^ MS = p sin nr = p r sin wV, and, moreover, on 

resolving OR or OR' into normal and tangential components (viz. 

along and perpendicular to ON), MS = p t ; p n = p cos nr, p^ => 9 

" imif QG + QH Px + Pv Pn + Pn 

p cos nr, and OM = 5 = 9 — ^ — 



MR 



= V(MG 2 + GK*) = y/ J (y ° / nT + ^} 

1R ^ x _ ^ 
GK _ 2p, 



MG v » ' 

cos 2xra = cos MIR =~ = p ± &.. 

MK ^ x — p y 



tan 2x« = tan 1STMR = 



MG - Pa —p/ 



EXAMPLES. 

1. Given OR = 17. OR' = 12. nr = n'r' = 37°. 
Solution. 

Px + Py _ P+p' 29 _ 

2 cos nr 2 cos o/ 

l^ZJh — V(329.G04025 — 204) =^11.207. 



ELLIPSE OF STRESS. 91 



2, 


Given OK = 12.471. OR' = 1.225. nr = ?iV = 42°. 


3. 


Given OR = 37.852. OP/ = 10.012. nr = n'r' = 67°. 


4. 


Given OK = 4.327. OP/ = 12.427, nr = nY = 48°. 


5. 


Given p* =10.327. p n r = 8.432. p t =p t ' =6.275 


6. 


Given p n =28.124. p B A =20.342. p t = Pt ' =3.427 


7. 


Given p n =16.327. ;;/ =12.437. ^ =;V = 6.538. 


8. 


Given p n =18.532. p n ' =15 428. p t = p( =4.836 




Problem V. This will be much better understood by taking t 



case in which it is applied, viz., earthwork and retaining wall 
Suppose we have a retaining wall, with a mass of earth piled up 
behind it. Now the only force which causes a stress is that of 
gravity, and this acts vertically ; hence, if we imagine a plane AB 
parallel to the upper surface, the stress (crashing) on this plane is 
due to the weight of the earth above, and nothing else ; hence it 
acts in a vertical line ; therefore by the proposition on conjugate 
stress, the stress on a vertical plane must be in a direction parallel 
to AB. These are two conjugate stresses. The last is confirmed 
by the actual state of the case, for if the wall were suddenly re- 
moved, the earth on top must necessarily begin to move in the 
direction CD. Now we know that if the obliquity of the stress is 
too great, that is, if the bank of earth is too steep it will fall, and 
there is a certain limiting angle of obliquity, which is the natural 
angle of rest different for different materials, and which is ascer- 
tained by experiment. This is called the angle of repose. 

Now the object of Problem Y, as applied to this particular case, 
would be, Given the slope of the earth, and also the angle of re- 
pose, to find the ratio between these two conjugate thrusts, that is, 
the ratio between the vertical stress on the oblique plane (parallel 
to the upper surface), and the oblique thrust on the vertical plane 
of the wall. When this is known we can easily ascertain the 
amount of pressure on the wall, for we know the vertical pressure 
when we know. the weight of the earth. Now let c be the angle 
of greatest obliquity of the stress, or the slope of the earth, 
p x + p y the intensities of the principal stresses (it makes no dif- 



92 ELLIPSE OF STRESS. 



ference where they are situated), and we have x ■ — - = sin <p 

Px ~t~ p y 

by Problem III. 

Now in Case 3, Problem IV, divide 20 by 19, and we have 

p x — p y I f 4pp' cos 2 nr \ 

p7+F = V t 1 ~ (p+p'Y ) 

I ( 4m/ cos 2 nr ) 4pp r cos 2 nr 
.*. sin co = 1 / i 1 — -7 j — jvqT r •'• sm <P — 1 — "~7 i — 7\sT 

V ( (p+pY ) (p+p) 2 

App' cos 2 nr (p + J?') 2 cos 2 nr 

.*. 1 — sin 2 o = cos 2 <p = — t j 7V2" .'• — i — t — = o — - 

(P + .P ) 4pjt/ cos 2 ^ 

(p ~\~p'y — ^pp r c ° s2 nr ~ c ° s2 ^ (p - p'y cos2 ^ r - cos 2 ^ 

*"" "" App f cos 2 c? ' \pp' cos 2 ^ 

(p — p r ) 2 cos 2 wr — cos 2 <p 

Divide (25) by this, and ~ ( — ■ / N2 = ^ 

v y - 7 ' \P \ P) cos 2 ?ir 

p — p r V(cos 2 nr — cos 2 <p) . . , ■ -. . 

.-. l r - L -, = ■, by composition and division. 

p -\- p cos nr J r 

»' cos nr — V (cos 2 nr — cos 2 ?>),... 

- = j >t o i — \i this is the required ratio. 

p cos nr + v ( cos nr — cos 9) 

When nr = then the stresses become principal stresses, and 

p Y 1 — </(l — cos 2 <p) 1 — sin ^ ,...', , 

— = r — j 77^ o — \ — 1 — i — ; 5 this is the case when 

jp x 1 + y(l — cos 2 ^) 1 + sm <p 

the upper surface is horizontal. 

v' 

When nr = <p, — = 1? that is, when the earth is piled up at 
the angle of repose, the vertical and oblique pressures are equal. 



EXAMPLES. 

1. Given <p = 35°, nr = 17°, find^-. 

p 

p f cos nr - \/(cos 2 nr - cos 2 <p) cos 17° - V (cos 2 17° - cos 2 35°) 

p cosnr + -s/(cos 2 nr-cos 2 ^) cos 17° + V (cos 2 17° - cos 2 35°) 

.9 5630- Vj(.95630) 2 - (.81915) 2 j _ .95630- .49346 _ .46284 

"".95630 + y^(.95630) 2 ~(.81915) 2 j "".95630 + .49346 "1.44976 

p' log .46284= L6654309 

— — 81.°) 25 

p —.ouzo. log 1.44976 = 0.1612961 

T.5041348 



EXAMPLES. 



93 



2. Given cp = 60°. nr = 25°. 

3. Given <p = 45°. nr — 0°. 

4. Given <p = 0°. nr = 0°. 

5. What is the pressure against the wall at the depth of 10 
feet, where w = 38°, nr = 16°, and 1 cubic foot of the earth 
weighs 120 lbs.? 

6. Solve the same example when <p = 72°, nr = 0°. 









EXAMPLES 


OX AETICLE 113. 




Given p 
np 


= 15, 

= 30°, 


12. 17, 18, 23. 

45°, 73°, 54°, 37°. 






Solution 










V 


np 


2np 


cos 2np 


sin 2np 


p cos 2np 


p sin 2np 


15 
12 
17 

18 
23 


30° 
45° 
73° 
54° 
37° 


60° 

90° 

146° 

108° 

74° 


.50000 

.00000 

—.82904 

—.30902 

.27564 


.86603 
1.00000 
.55919 
.95106 
.96126 


7.50000 

0.00000 

—14.09368 

—5.56236 

. 6.33972 


12.99045 

12.00000 

9.50623 

17.11908 

22.10898 


85 


—5.81632 


73.72474 






P- + Py _ . „ _ 
2 2 " 


[2.5. 




P* — Py _ 

2 


= h*S\z 


3.8145783424 + 543.: 
jo x = 116.453. Pj = 


3.3372880676^ 
—31.453. 


= 73.953 




tan 


X p sin 2np 

2nx = v . = 

2. p COS Z7ZJ0 


73.72474 

5.816.32 




log 73.7 


2474 


= 1.8676132 






log 5i 


$1632 


= 0,7646483 






log tan 


(180 — 


2nx) = 1.1029649 






1 


B0 — 2 


raa: = 85° 29' 21". 2 
nx= 47° 15' IS 


nx = 04° 30' 


39'. 


2. Gii 


T en 


3. Given 4. 


Given 


o. Given 


1 
1 
1 
] 


V rq 
43 
4 72 
7 12 
6 17 
>7 '85 


) 

o 
o 

o 
o 
o 


P n l 
1.125 30 
3.427 45 
5.642 60 
3.721 15 
8.347 75 


o p 
4 
12 
19 
32 
73 


np 

10° 

57° 

153° 

149° 

190° 


p np 
8 14° 
7 17° 
3 81° 
2 94° 
1 73° 



94 EQUILIBRIUM OF FLUIDS. 

Articles 114, 115, 116 and 117, of Rankine should now be care- 
fully studied. 

Note on Art. 116. 

It is important, in this connection, to keep clearly in mind the 
meaning of the symbols p xx , etc., hence I repeat them here, viz.? 
jt? xx means the intensity of the stress on a plane perpendicular to 
Ox in the direction Ox. 

So p xy means the intensity of the stress in the direction Oy, on 
a plane perpendicular to Ox ; and this we have already seen is 
equal to p yx , or the intensity of the stress in the direction Ox, on 
a plane perpendicular to Oy. 

Secondly. 4- represents the rate of variation of the stress 

along Oz, that is the amount of increase of stress for a unit of dis- 

dp 
tance, and hence -,- Az will be the amount of increase for the dis- 

dp 
tance Az, or the whole intensity will be p + ~r Az. 

Thirdly. The equations 1 are obtained by imposing the three 
known conditions of equilibrium, viz., 

Z P cos a = 0, 2 P cos (S = 0, - V P cos y = 0; 
that is, by placing the resultant stress in the direction of each axis 
equal to 0. 

ANOTHER VIEW OF ARTICLE 117. 

Suppose we have a body of liquid confined in a vessel of any 
shape ; the free upper level will, as we already know, be horizon- 
tal, and the pressure at any point is due merely to the weight of 
the water above it. 

•The fact that any molecule, as EFGH, is at rest, shows us that 
the horizontal pressures acting on the molecule are in equilibrium. 

Suppose the intensity of the pressure on the plane AB be p, the 
increase of pressure on the plane CD is equal to the weight of the 
liquid included between these two planes, AB and CD, thus the 



EXAMPLES. 95 

whole amount of increase is ivzlz- surf. CD, where w is the weight of 

• . wzlx surf. CD 

a unit of volume, and Az = EG, hence its intensity is — f q-q — 

= iv J #, or -r J a; ; hence the total intensity of the stress on the 

dp 
plane CD is p + w J x, or p + -^/J x. 

If, therefore, jd is the intensity of the pressure at the free upper 

r*dp 
level, the intensity p at any depth is equal to p + J q ^r tte = 

j) -f zrefo, which, when w is the same at all depths, becomes 

P= z Po + wx. 



EXAMPLES. 



1. What is the intensity of the pressure of the water contained 
in a cask, at a depth of 12 feet below the free upper level. 

Solution, p — p + icx. And p = 14.7 lbs. w = 6 t 7 W" lbs. 
^=12X12 = 144 inches. 

.-.p = 14.7 + &**ff^i±± = 14.7 + 5.2 = 19.9 lbs. 

This is the pressure at the given depth, but if we were deter- 
mining the pressure of the water on the sides or bottom of the cask 
at that depth, we should make a gross error by the solution, for this 
pressure is counteracted in part by the pressure of the atmosphere 
outside, and hence the resultant pressure against the sides of the 
cask is only 5.2 lbs. per square inch, for this case j? = 0, and .\ 
p '= iv x. 

2. The free upper level of a reservoir is situated at a height of 
80 ft. above the level of the ground ; what is the intensity of the 
pressure on a pipe leading from the reservoir at a height of 10 ft. 
above the level of the ground ? 

3. What is the intensity of the pressure at the foot of a mill- 
dam 30 ft. high, the free upper level of the water being 1 ft. above 
the top of the dam ? 

4. What is the total pressure of water on the vertical face of a 
reservoir wall 10 ft. high, where the free upper level is even with 
the top of the wall ? 



9G FLOATING BOD1LS. 

5. What is the total normal pressure, and what its mean in- 
tensity, on a reservoir wall whose back makes with the horizontal* 
an angle of 75° ? Also find its intensity at the foot of the wall. 

0. Given the following reservoir walls with an inclined face. 

1. height = 10. 2. height = 15. 3. height = 14. 

a = 60°. a = 58° 3' a = 83°. 

A SHORTER VIEW OF ARTICLE 120. 

EQUILIBRIUM OF A FLOATING BODY. 

In the case of a floating body, as, for instance, a ship which is 
partly immersed in water, and thus is in equilibrium, if w r e im- 
agine the body removed, its place would be filled with water, and this 
water is kept in equilibrium by the resultant pressure of the water 
around it ; but to support it it requires a vertical force equal to the 
weight of this body of water, and acting vertically upwards, and so 
likewise to support the floating body we must have a resultant up- 
ward pressure equal to the weight of the body. Hence 

A floating body displaces a volume of water whose weight is 
equal to the weight of the body itself. 

The centre of gravity of the liquid displaced is called the Centre 
of Buoyancy of the body. Now the weight of the body acts 
vertically downward through the centre of gravity ; the upward 
pressure of the water acts vertically upwards through the centre of 
buoyancy ; hence for equilibrium we must have these two in the 
same vertical line. 

For Article 121 of Rankine a similar reasoning holds. 

APPARENT WEIGHTS. 

We have already seen that a body immersed in a liquid is 
pressed upon by a force whose resultant acts vertically upward, 
and equal in amount to the weight of an equal volume of the 
liquid ; so also a body in the air is pressed upwards by a force 
equal to the weight of an equal volume of air. 



IMMERSED PLANE. 97 

Thus if we have two bodies, whose absolute weights are equal, 
i. e., which contain the same amount of matter, and on which, con- 
sequently, gravity acts equally, suspended one in each pan of a 
balance of equal arms, if the bodies are not equal in size the air 
presses unequally on them, and hence their weights are diminished 
by unequal amounts, and do not therefore appear equal. 

The weight of a given volume of air varies directly as the 
pressure. 

Also the volume of any given weight of air is increased by 
46V.2 f° r eacn degree of temperature, through which it is raised 
from 0° Fahrenheit upwards. Hence let v denote the volume oc- 
cupied by an amount of air whose weight at a temperature of 0° 
Fahrenheit, is w; when the temperature is raised to t° . the volume 



will be t 



( l + 46T72> 



Let v° denote the volume occupied at 0° Fahrenheit, by a weight 
w of air, then when the temperature is raised to 32° its volume 

/ 32 \ . 

will be v Q I 1 -j— TiTTT) )* and if each of these last volumes is taken 

to be one cubic foot, we have 

v 461 — |— 32 w v Q 

v 4bl -f- t w Q v 

w 493 493 



w = W* 



* * w ~~ 461 + t " u ~ w * 461.2 + t 
This determines w when the pressure is not changed, but if we 
have a pressure of p atmospheres, the formula becomes 

493 

W= ^461.2 + ; 

PRESSURE ON AN IMMERSED PLANE. 

Let AD (Fig. 57) represent a section of the free upper level, 
made by a vertical plane, and BC, that of the inclined plane, the 

13 



98 



TMMERSED PLANK. 



vertical plane being perpendicular to both ; then we know that the 
pressure at any point F is equal to w FG multiplied by the area 
pressed upon ; hence if we divide the inclined plane into small rec- 
tangles, and call the area of each one r, we shall have for the pres- 
sure on any one of them wxr, and hence the total pressure P = 
2wxr = w£xr. But we know £xr = x {) Er, where x$ is the coor- 
dinate of the centre of gravity of the surface, hence P = wx £r 



= wx Area BC, 



Po 



-WXy. 



AreaBC 

Art. 125 of Rankine has already been discussed in the problems 
on earthwork. 









wf^^m^ - 



' A? f 'Wi A jsi' 



v.r-'vwv* ^ 









..-. :,,,„: 



a A _ /- , ' , i-> /s ** ->. ' ■ * '.* r\ f\ r\ 









. fi:'^^: 



SSI ft 









A>M»A' ■■' a*» „.»« 



\*A Aft 






»AAA* 



;AM?V/ 






* 



^mmKf^m 



- aa;>a"^'P^ 



- »i 












M^W^9^ 









\r\f\rsf\i\* 






K * i 

J A r.r\ f 















"A 















Vtf^ 






LIBRARY OF CONGRESS 

II! II II 



in mi mi ii 



I 



IliliilllilllllMIIIIIlJlll 
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